- (a) Suppose that $V$ is a finite-dimensional vector space over a field $𝔽$, and that $T: V→V$ is a linear transformation.
(i) Prove that there exists a non-zero polynomial $p(x)$ such that $p(T)=0$.
(ii) Prove that there exists a unique monic polynomial $m(x)$ such that for all polynomials $q(x), q(T)=0$ if and only if $m(x)$ divides $q(x)$.
(iii) State a criterion for diagonalisability of $T$ in terms of $m(x)$.
(b) Suppose that $V$ is a finite-dimensional vector space over a field 𝔽 and $T: V →V$ is a linear transformation.
(i) Prove that for all $i,\ker T^i$ is a subspace of $\ker T^{i+1}$.
Let $B_1 \subseteq B_2 \subseteq \cdots$ be sets such that $B_i$ is a basis for $\ker T^i$.
(ii) Deduce that if for some $k, T^k=0$, then $T$ is upper-triangularisable. Deduce that for any $\lambda∈𝔽$, if $(T-\lambda I)^k=0$, then $T$ is upper-triangularisable.
(iii) Show that $T$ is upper-triangularisable if and only if $m(x)$ is a product of linear factors. [You may use the Primary Decomposition Theorem.]
(c) For which values of $α$ and $\beta$ is the matrix
\[ A=\left(\begin{array}{ccc} 2 & 1 & -1 \\ α-1 & α-\beta & \beta \\ α-1 & α-\beta-1 & \beta+1 \end{array}\right) \] diagonalisable over $ℝ$ ?
For which values of $α$ and $\beta$ is it upper-triangularisable over $ℝ$ ?
(a) (i) If the dimension of $V$ is $n$, then that of $\operatorname{Hom}(V)$ is $n^2$. So, $\left\{I, T, T^2, \ldots, T^{n^2}\right\}$ is linearly dependent. Hence there exist constants $α_0, α_1, \ldots, α_{n^2}$ not all zero such that \[ \sum_{i=0}^{n^2} α_i T^i=0 . \] Let $p(x)=\sum_{i=0}^{n^2} α_i x^i$
Then $p(x)$ is a non-zero polynomial and $p(T)=0$, as required.
(ii) Let $p(x)$ be a non-zero polynomial of minimal degree such that $p(T)=0$.
Define $m(x)$ to be the result of dividing $p(x)$ by its leading coefficient.
Then $m(x)$ is monic, and $m(T)=0$. Hence if $m(x)$ divides $q(x)$, then $q(T)=0$.
Now suppose that $q(x)$ is a polynomial such that $q(T)=0$.
Then there exist polynomials $a(x)$ and $b(x)$ such that
\[
q(x)=a(x) m(x)+b(x),
\]
and $b(x)=0$ or the degree of $b(x)$ is less than that of $m(x)$.
But then $b(x)$ must be zero, yielding that $m(x)$ divides $q(x)$, for otherwise $b(x)=q(x)-a(x) m(x)$, so $b(T)=0$, contradicting the minimality of the degree of $m(x)$.
(iii) $T$ is diagonalisable if and only if $m(x)$ is a product of distinct linear factors.
(b) (i) Suppose that $v \in\ker T^i$.
Then $T^i(v)=0$.
Hence $T\left(T^i(v)\right)=0$.
That is, $T^{i+1}(v)=0$.
So $v \in\ker T^{i+1}$.
(ii) Writing $B_k$ with the elements of $B_1$ first, followed by the elements of $B_2 \backslash B_1$, and so on, the matrix of $T$ with respect to $B_k$ is upper-triangular, and indeed all diagonal entries are zero.
We justify the statement that the matrix is upper-triangularisable as follows. The first few columns correspond to element of $B_1$, which belong to the kernel of $T$, and so have no non-zero entries at all. Any subsequent column corresponds to an element of $\ker T^{i+1} \backslash\ker T^i$, for some $i$, which is sent by $T$ to an element of ker $T^i$, so to a linear combination of members of the basis which are strictly earlier in the ordering. So all non-zero entries in that column are strictly above the diagonal.
If $(T-\lambda I)^k=0$, let $S=T-\lambda I$. Then $S$ is upper-triangularisable. It follows immediately that $T$ is.
(iii) Now suppose that \[ m(x)=\prod_{i=1}^r\left(x-\lambda_i\right)^{k_i} \] By the Primary Decomposition Theorem,$$V=\bigoplus_{i=1}^{r} \ker\left(T-\lambda_i I\right)^{k_{i}}$$ Let $B_i$ be a basis of $\ker\left(T-\lambda_i I\right)^{k_i}$ with respect to which $T↾_{\ker\left(T-\lambda_i I\right)^{k_i}}$ is upper-triangularisable. Then if $B=\bigcup_{i=1}^r B_i$, then the matrix of $T$ with respect to $B$ is upper-triangular.
That $m(x)$ splits into linear factors if $T$ is upper-triangularisable is obvious; because if $\lambda_1, \ldots, \lambda_n$ are the diagonal entries, then $\prod_{i=1}^n\left(T-\lambda_i I\right)$ is strictly upper-triangular (that is, all diagonal entries are zero), and therefore idempotent. Thus for some $k$ (in fact, for some $k \leqslant n$ ), $\left(\prod_{i=1}^n\left(T-\lambda_i I\right)\right)^k=0$. Thus $m_T(x)$ divides $\left(\prod_{i=1}^n\left(x-\lambda_i\right)\right)^n$, and thus splits into linear factors.
(c) Let \[ A=\left(\begin{array}{ccc} 2 & 1 & -1 \\ α-1 & α-\beta & \beta \\ α-1 & α-\beta-1 & \beta+1 \end{array}\right) . \] Then \begin{aligned} \operatorname{det}(A-x I)=&(2-x)((α-\beta-x)(\beta+1-x)-\beta(α-\beta-1)) \\ &-((α-1)(\beta+1-x)-\beta(α-1)) \\ &-((α-1)(α-\beta-1)-(α-1)(α-\beta-x)) \\ =&(2-x)((α-\beta-x)(\beta+1-x)-\beta(α-\beta-1)) \\ &+(1-α)((\beta+1-x)-\beta+(α-\beta-1)-(α-\beta-x)) \\ =&(2-x)\left(x^2+x(-1-α)+α\right) \\ =&(2-x)(x-1)(x-α) . \end{aligned} If $α$ is not equal to 1 or 2 , then $\chi_A(x)$ has three distinct roots and so $A$ is diagonalisable.
If $α$ is equal to 1 or 2 , then $\chi_A(x)$ has a repeated root, and $A$ is diagonalisable if and only if $(A-I)(A-2 I)=0$.
Now the $(2,1)$-entry of $(A-I)(A-2 I)$ is $(α-1)^2$, which is not zero unless $α=1$. So if $α=2$, $A$ is not diagonalisable.
If $α=1$, then \[ (A-I)(A-2 I)=\left(\begin{array}{ccc} 1 & 1 & -1 \\ 0 & -\beta & \beta \\ 0 & -\beta & \beta \end{array}\right)\left(\begin{array}{ccc} 0 & 1 & -1 \\ 0 & -\beta-1 & \beta \\ 0 & -\beta & \beta-1 \end{array}\right)=\left(\begin{array}{ccc} 0 & 0 & 0 \\ 0 & \beta & -\beta \\ 0 & \beta & -\beta \end{array}\right) \] which is zero if and only if $\beta=0$.
So $A$ is diagonalisable if and only if either $α$ is not 1 or 2 , or $α=1$ and $\beta=0$.
By the criterion in part (b), $A$ is upper-triangularisable whatever the values of $α$ and $\beta$, since by the Cayley-Hamilton Theorem $m(x)$ divides $(x-2)(x-1)(x-α)$ and so is a product of linear factors. - (a) Suppose that $V$ is a finite-dimensional vector space over a field $𝔽$. Suppose that $B=\left\{e_1, \ldots, e_n\right\}$ is a basis for $V$.
(i) Define the dual space $V'$ of $V$ and the dual basis $B'=\left\{e_1', \ldots, e_n'\right\}$. Prove that $B'$ is indeed a basis for $V'$.
(ii) If $T: V→V$ is a linear transformation, define the dual map $T'$. State and prove a relationship between the matrices of $T$ and $T'$ with respect to the bases given. How are the characteristic polynomials of $T$ and $T'$ related? How are the minimum polynomials related? Justify your answers briefly.
(iii) If $U$ is a subspace of $V$, define the annihilator $U^∘$ of $U$.
(iv) Define a natural isomorphism $Φ$ between $V$ and its double dual $V''$. (You do not need to give proofs that $Φ$ is well-defined or that it is an isomorphism.) Prove that if $U$ is a subspace of $V$, then $\left.Φ\right|_U$ is a bijection between $U$ and $U^{∘∘}$.
(b) Let $V$ be the vector space of all functions $f: \mathbb{N}→ℝ$ such that for all but finitely many $n, f(n)=0$, equipped with operations of vector addition and scalar multiplication defined so that $(f+g)(n)=f(n)+g(n)$ and $(α f)(n)=α f(n)$ for all $f, g∈V$, $n∈\mathbb{N}$, and $α∈ℝ$.
Define $W$ to be the vector space of all functions from $\mathbb{N}$ to $ℝ$, with similarly defined operations of vector addition and scalar multiplication.
If $f∈W$, define $\theta_f: V→ℝ$ so that \[ \theta_f(g)=\sum_{n=0}^{\infty} f(n) g(n) \] Prove that the map $f \mapsto \theta_f$ is an isomorphism between $W$ and $V'$.
Prove that the map $Φ: V→V''$ defined as in part (a) is not a surjection.
[You may assume that if $U$ is a vector space over ℝ, $L$ is a linearly independent subset of $U$, and $h: L→ℝ$, then there exists a linear functional $k: U→ℝ$ such that $\left.k\right|_L=h$]
(a) (i) The dual space $V'$ is the set of all linear functionals on $V$, that is to say, the set of all functions $f: V→𝔽$ such that $f(u+v)=f(u)+f(v)$ and $f(α v)=α f(v)$ for all $α∈𝔽$ and all $u, v∈V$, with vector addition and scalar multiplication defined so that $(f+g)(v)=f(v)+g(v)$ and $(α f)(v)=α f(v)$ for all $v∈V, f, g∈V'$ and $α∈𝔽$.
The dual basis is defined so that $e_i'\left(e_j\right)=\delta_{i, j}$.
The dual basis is linearly independent, since if \[ α_1 e_1'+\cdots+α_n e_n'=0, \] then for all $i$, \[ \left(α_1 e_1'+\cdots+α_n e_n'\right)\left(e_i\right)=0, \] that is, $α_i=0$.
To prove that it is a spanning set, suppose that $f∈V'$. Let $α_i=f\left(e_i\right)$ for all $i$. Then for all $i$, \[ f\left(e_i\right)=α_i=\left(\sum_j α_j e_j'\right) e_i \] so since $f$ and $\sum_j α_j e_j'$ are linear and agree on a spanning set, they are equal.
(ii) If $f∈V'$, then define $T'(f)$ so that $T'(f)(v)=f(T(v))$ for all $v∈V$.
Let the matrix of $T$ with respect to $B$ be $\left(a_{i, j}\right)$ and the matrix of $T'$ with respect to $B'$ be $\left(b_{i, j}\right)$. Then\[e_i'\left(T\left(e_j\right)\right)=e_i'\left(\sum_{k=1}^n a_{k, j} e_k\right)=a_{i, j}\]while\[\left(T'\left(e_i'\right)\right)\left(e_j\right)=\left(\sum_{k=1}^n b_{k, i} e_k'\right)\left(e_j\right)=b_{j, i} .\]So $b_{j, i}=a_{i, j}$, and the matrices are each other's transpose; and so their minimum polynomials are the same, as are their characteristic polynomials.
(iii) $U^∘=\left\{f∈V'∣\forall u∈U: f(u)=0\right\}$.
(iv) $Φ$ is defined so that for all $f∈V'$ and $v∈V$,\[Φ(v)(f)=f(v) .\]We show that $u∈U$ if and only if for all $f∈U^∘, f(u)=0$. The forward direction is simply the definition of $U^∘$. As for the reverse direction, let $\left\{e_1, \ldots, e_k\right\}$ be a basis for $U$ and extend it to a basis $\left\{e_1, \ldots, e_n\right\}$ for $V$. Let $\left\{e_1', \ldots, e_n'\right\}$ be the dual basis. Then $\left(\sum_{j=1}^n α_j e_j'\right)\left(e_i\right)=0$ if and only if $α_i=0$. It follows that $f\left(e_i\right)=0$ for all $i≤k$ if and only if $f$ is in the span of $\left\{e_{k+1}', \ldots, e_n'\right\}$. It now readily follows that $U^∘$ is the span of $\left\{e_{k+1}', \ldots, e_n'\right\}$.
Now, $u∈U$ if and only if for all $f∈U^∘, f(u)=0$, if and only if for all $f∈U^∘, Φ(u)(f)=0$, if and only if $Φ(u)∈U^{∘∘}$.
(b) If $f∈W$, we observe that $\theta_f$ is linear, so is an element of $V'$. Also, if $f \neq 0$, then there exists $n∈ℕ$ such that $f(n) \neq 0$. Now we define $g∈V$ such that $g(n)=1$, and $g(m)=0$ for all $m \neq n$. Then $\theta_f(g)=f(n) \neq 0$. So the operator $f \mapsto \theta_f$ is one-to-one. Finally, to show that it is onto, let $h$ be any element of $V'$. Then if $g_n$ is defined, for each natural number $n$, so that $g_n(m)=1$ if $m=n$ and is equal to 0 otherwise, then the set of $g_n$ is a basis for $V$. So if $f$ is defined so that $f(n)=h\left(g_n\right)$ for each $n$, then for any $g∈V, g=\sum_n g(n) g_n$, and $\theta_f(g)=\sum_n f(n) g(n)=\sum_n h\left(g_n\right) g(n)=h\left(\sum_n g(n) g_n\right)=h(g)$. So $h=\theta_f$.
For each $n$, define $f_n(m)$ to be 1 if $n=m$ and 0 if $n \neq m$. Let $g$ be the function $n \mapsto 1$. Then $\left\{f_n: n∈\mathbb{N}\right\} \cup\{g\}$ is linearly independent in $W$, and so its image under the operator $f \mapsto \theta_f$ is linearly independent in $V'$.
Define $h\left(\theta_{f_n}\right)$ to be 0 and $h(g)$ to be 1. Extend this to a linear functional $k$ on $V'$.
Since $k\left(\theta_{f_n}\right)=0$ for all $n$ and $k(g)=1$, $k$ cannot be in the image of $Φ$. - Let $V$ be a finite-dimensional inner-product space over $\mathbb{C}$.
(a) Suppose that $T: V→V$ is a linear transformation. Define the adjoint map $T^*$.
Suppose that $T$ has the property that $T^*=α T$ for some $α∈\mathbb{C}$. Prove that $T$ is diagonalisable.
(b) We say that $T$ is self-adjoint if $T^*=T$, and that it is skew-adjoint if $T^*=-T$. Observe that if $S$ and $T$ are self-adjoint, then so are $S+T, S-T$, and $\beta T$, for any real number $\beta$.
Recall that if $T: V→V$ is any linear transformation, then $T+T^*$ is self-adjoint.
(i) Prove that any linear transformation $T$ can be written as the sum of a self-adjoint and a skew-adjoint linear transformation.
Is it the case that a sum of diagonalisable linear transformations is diagonalisable? Give a proof or a counterexample.
(ii) What are the possible eigenvalues of a self-adjoint linear transformation? Justify your answer carefully.
(iii) Characterise the possible Jordan Normal Forms of linear transformations $T: V→V$ such that $T^2$ is self-adjoint.
(c) Suppose now that $T: V→V$ is a linear transformation, and that $T T^*=T^* T$.
(i) Prove that if $v$ is an eigenvector of $T^*$, then $v^{\perp}$ is $T$-invariant.
(ii) Prove that if $V_\lambda=\operatorname{ker}(T-\lambda I)$, and $v∈V_\lambda$, then $T^* v∈V_\lambda$ also.
(iii) Hence prove that there exists an orthogonal basis for $V$ consisting of vectors which are eigenvectors for both $T$ and $T^*$.
(iv) Does it follow that $T$ is self-adjoint? Give a proof or a counterexample.
(a) The adjoint is the unique linear transformation $T^*: V→V$ such that for all $u, v∈V$, $\left(T^* v, u\right)=(v, T u)$.
Suppose that $T^*=α T$, where $α \neq 0$. Assume that $V$ is not trivial. Since the underlying field is $\mathbb{C}, \chi_T(x)$ has a root, so $T$ has an eigenvector, $v$; say $\lambda$ is the eigenvalue.
We prove that $v^{\perp}$ is $T$-invariant.
Suppose that $u∈v^{\perp}$.
Then $(u, v)=0$.
Also $(T u, v)=(\lambda u, v)=\lambda(u, v)=0$.
So $\left(u, T^* v\right)=0$.
Now $T^* v=α T v$, so $(u, α T v)=0$, so $\bar{α}(u, T v)=0$, so since $α \neq 0,(u, T v)=0$ as required.
By the inductive hypothesis we assume that $T↾_{v^{\perp}}$ has a basis $B$ of eigenvectors. Then $B \cup\{v\}$ is a basis of eigenvectors for $T$.
(b) (i) $T-T^*$ is clearly skew-self-adjoint.
$T=\frac12\left(T+T^*\right)+\frac12\left(T-T^*\right)$ as required.
The linear transformation with matrix with respect to the standard basis given by \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} is not diagonalisable, since its characteristic polynomial is $x^2$ and its minimum polynomial is not $x$.
But it is the sum of a self-adjoint and a skew-self-adjoint transformation as above.
(ii) $I$ is certainly self-adjoint so for all real $\beta, \beta I$ is self-adjoint also, and has eigenvalue $\beta$.
Conversely, if $T$ is self-adjoint with eigenvalue $\lambda$, then $(T v, v)=(\lambda v, v)=\lambda\|v\|^2$, while $(v, T v)=(v, \lambda v)=\bar{\lambda}\|v\|^2$, so $\lambda=\bar{\lambda}$ and $\lambda$ is real.
(iii) Suppose that $T^2$ is self-adjoint and \[ A=\left(\begin{array}{cccc} \lambda & 1 & \ldots & 0 \\ 0 & \lambda & & \\ & & & \lambda \end{array}\right) \] is a Jordan block for $T$.
Then $A^2$ has the form \[ \left(\begin{array}{ccccc} \lambda^2 & 2 \lambda & 1 & … &…& 0 \\ 0 & \lambda^2 & 2 \lambda & ⋱\\ & &⋱ & ⋱&⋱\\ & & &\lambda^2 & 2\lambda&1\\ & & &&\lambda^2 & 2\lambda\\ & & & && \lambda^2 \end{array}\right) \] and is diagonal if and only if either the size of the block is $1 \times 1$, or it has size $2 \times 2$ and $\lambda=0$.
Also, $A^2$ is diagonalisable if and only if it is diagonal; for if it is not diagonal then its minimum polynomial is $\left(x-\lambda^2\right)^k$ for some $k>1$, which is not a product of distinct linear factors.
So the Jordan Normal Forms of transformations $T$ such that $T^2$ is self-adjoint have Jordan blocks of that form, with $\lambda$ being either real or purely imaginary.
(c) (i) Suppose $v$ is an eigenvector of $T^*$, and $u∈v^{\perp}$.
Then $(v, u)=0$.
Since $T^* v$ is a scalar multiple of $v,\left(T^* v, u\right)=0$.
Hence $(v, T u)=0$, and so $T u∈v^{\perp}$, as required.
(ii) Suppose that $v∈V_\lambda$.
Then $T^* T v=T^*(\lambda v)=\lambda T^* v$. But also $T^* T v=T T^* v$. Hence $T\left(T^* v\right)=\lambda T^* v$, and so $T^* v∈V_\lambda$
Schur's lemma
(iii) If $V$ is non-trivial, then the characteristic polynomial of $T$, being a non-constant complex polynomial, has a root. So $T$ has an eigenvalue $\lambda$, whose corresponding eigenspace $V_\lambda$ is nontrivial. Now $\left.T^*\right|_{V_\lambda}$ also has an eigenvector by the same reasoning, which is a simultaneous eigenvector of $T$ and $T^*$.
We do induction on $\dim V$.
Let $u$ be a simultaneous eigenvector for $T$ and $T^*$. Then $u^{\perp}$ is invariant under both $T^*$ and $T$. By the inductive hypothesis, $u^{\perp}$ has a basis $B$ of the correct form.
Then $B \cup\{u\}$ is a basis of the desired form for $V$.
(iv) If $T=\mathrm{i} I$, then $T^*=-\mathrm{i} I$. These commute, but are not equal.Normal matrix
Among complex matrices, all unitary, Hermitian, and skew-Hermitian matrices are normal, with all eigenvalues being unit modulus, real, and imaginary, respectively. Likewise, among real matrices, all orthogonal, symmetric, and skew-symmetric matrices are normal, with all eigenvalues being complex conjugate pairs on the unit circle, real, and imaginary, respectively. However, it is not the case that all normal matrices are either unitary or (skew-)Hermitian, as their eigenvalues can be any complex number, in general. For example, \[A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}\] is neither unitary, Hermitian, nor skew-Hermitian, because its eigenvalues are $2, (1\pm i\sqrt{3})/2$; yet it is normal because \[AA^* = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix} = A^*A.\] For the curious, the four classes \begin{bmatrix} a & b & 0 \\ 0 & a & b \\ b & 0 & a \end{bmatrix} \begin{bmatrix} a & b & 0 \\ 0 & a & -b \\ b & 0 & a \end{bmatrix} \begin{bmatrix} a & b & 0 \\ 0 & a & b \\ -b & 0 & a \end{bmatrix} \begin{bmatrix} a & b & 0 \\ 0 & a & -b \\ -b & 0 & a \end{bmatrix} are neither unitary nor skew-Hermitian for all non-zero real $a$ and $b$. There are more 3×3 examples, but among 2×2 matrices, there are only ones that are multiples of unitary matrices.