# Ph.D. Comprehensive exam practice problems, Round 2

Exercise 1 Let ${V}$ be the vector space of continuous real-valued functions on the interval ${[0,\pi]}$. Then, for any ${f \in V}$,

$\displaystyle 2 \int_0^\pi f(x)^2 \sin x dx \geq \left(\int_0^\pi f(x) \sin x dx\right)^2.$

Proof: Let ${d\mu}$ be the measure ${\frac{\sin x dx}{2}}$ on ${[0,\pi] = X}$. Then ${(X, d\mu)}$ is a probability space, ${f}$ is Lebesgue-integrable on ${X}$ and ${t \mapsto t^2}$ is a convex function ${\mathbf R \rightarrow \mathbf R}$. By Jensen’s inequality,

$\displaystyle \int_0^\pi f(x)^2 d\mu \geq \left(\int_0^\pi f(x) d\mu\right)^2.$

Multiplying throughout by ${4}$ we get the claimed inequality.
$\Box$

Exercise 2 Let ${T}$ be a linear operator on a finite-dimensional vector space ${V}$. (a) Prove that if every one-dimensional subspace of ${V}$ is ${T}$-invariant, then ${T}$ is a scalar multiple of the identity operator. (b) Prove that if every codimension-one subspace of ${V}$ is ${T}$-invariant, then ${T}$ is a scalar multiple of the identity operator.

Proof: (a) The hypothesis means that every nonzero vector of ${V}$ is an eigenvector of ${T}$. Suppose ${v_1, v_2}$ are eigenvectors of ${T}$ with eigenvalues ${\lambda_1}$, ${\lambda_2}$. Since, by assumption ${v_1+v_2}$ is also an eigenvector, and ${v_1}$ and ${v_2}$ are independent, we can read off the eigenvalue of ${v_1 + v_2}$ off of either coefficient in the equation ${T(v_1+v+2)= \lambda_1 v_1 + \lambda_2 v_2}$, and therefore ${\lambda_1 = \lambda _2}$. Therefore ${T}$ is a multiple of the identity operator.

(b) Let ${T^\vee}$ be the dual operator on ${V^\vee}$. We claim that ${T^\vee}$ satisfies the condition of ${(a)}$. First, we have the following:

Lemma 1 Two functionals ${f, g : V \rightarrow k}$ (where ${k}$ is the ground field) have the same kernel if and only if they are multiples of each other.

Proof: Indeed, it is trivial if either of ${f}$ or ${g}$ is ${0}$ (in which case both are zero), so suppose neither is ${0}$. Recall that if ${W \subseteq V^\vee}$ and we define ${\mathrm{Ann}(W) = \{v \in V : f(v) = 0\: \: \forall w \in W\}}$, then we have a canonical isomorphism ${\mathrm{Ann}(W) \cong (V^\vee/W)^\vee}$, which in particular implies ${\dim \mathrm{Ann}(W) = \mathrm{codim}(W\subseteq V^\vee)}$. If we apply this to ${W=\left}$, we have, under assumption,

$\displaystyle \mathrm{Ann}(W) = \ker f \cap \ker g = \ker f = \ker g$

which has codimension ${1}$ since ${f,g \neq 0}$. Therefore ${W}$ has dimension ${1}$, and ${f}$ and ${g}$ are scalar multiples of each other. $\Box$

Now, back to ${(b)}$. S suppose that ${0 \neq f \in V^\vee}$. Then ${\ker f}$ has codimension ${1}$ in ${V}$, and therefore, under the hypothesis of (b), ${T(\ker f) \subseteq \ker f}$. This implies ${\ker T^\vee(f) \supseteq \ker f}$; indeed, if ${v \in \ker f}$, then ${T^\vee(f)(v) = f(Tv) = 0 }$ since ${Tv \in \ker f}$. Since ${\ker f}$ is codimension ${1}$, we either have equality, or ${T^\vee(f) = 0}$. If there is equality, then ${T^\vee(f)}$ and ${\ker f}$ have the same kernel and therefore they are proportional, i.e. ${f}$ is an eigenvector of ${T^\vee}$. If ${T^\vee(f)=0}$ then ${f}$ is trivially an eigenvector of ${T^\vee}$. In every case, we see that ${f}$ is an eigenvector of ${T^\vee}$. By (a), ${T^\vee}$, and therefore ${T}$, is a multiple of the identity operator. $\Box$

Exercise 3 Let ${T}$ be a linear operator on a finite-dimensional inner product space ${V}$.

• (a) Define what is meant by the adjoint ${T^*}$ of ${T}$.
• (b) Prove that ${\ker T^* = \mathrm{im}(T)^\perp}$.
• (c) If ${T}$ is normal, prove that ${\ker T = \ker T^*}$. Give an example when the equality fails (and, of course, ${T}$ is not normal).

Proof:

• (a) It is the unique linear operator ${T^*}$ on ${V}$ such that ${\left = \left}$ for every ${v, w \in V}$.
• (b) Indeed,

$\displaystyle \begin{array}{rcl} v \in \ker T^* &\Leftrightarrow& \left = 0 \: \forall w \in W \\ &\Leftrightarrow& \left = 0 \: \forall w \in W \\ &\Leftrightarrow& v \perp T(w)\: \forall w \in W. \end{array}$

• (c) A normal operator is one which commutes with its adjoint, i.e. ${TT^* = T^*T}$. Thus,

$\displaystyle \begin{array}{rcl} v \in \ker T^* &\Leftrightarrow& \left = 0\\ &\Leftrightarrow& \left = 0 \\ &\Leftrightarrow& \left = 0 \\ &\Leftrightarrow& \left = 0\\ &\Leftrightarrow& Tv=0. \end{array}$

An example where the equality fails is supplied by the operator ${T=\left(\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right)}$ acting on ${(\mathbf R^2, \bullet)}$ in the standard way. The vector ${(1,-1)}$ is in the kernel of ${T}$ but not of ${T^*}$.

$\Box$

# Ph.D. Comprehensive exam practice problems, Round 1

In May, I will be taking the qualifying exams for my Ph.D.. Over the next few weeks, I will be posting practice problems and my solutions to them. Until the end of February, I will be reviewing linear algebra, single variable real analysis, complex analysis and multivariable calculus. In March and April, I will be focusing on algebra, geometry and topology.

Here are three problems to start.

Problem: Suppose that ${A}$ is an ${n \times n}$ real matrix with ${n}$ distinct real eigenvalues. Show that ${A}$ can be written in the form ${\sum_{j=1}^n \lambda_j I_j}$ where each ${\lambda_j}$ is a real number and the ${I_j}$ are ${n\times n}$ real matrices with ${\sum_{j=1}^n I_j = I}$, and ${I_jI_l = 0}$ if ${j \neq l}$. Give a ${2 \times 2}$ real matrix ${A}$ for which such a decomposition is not possible and justify your answer.

Solution: for each ${j}$, let ${E_j}$ denote the matrix with a ${1}$ on the entry ${(j,j)}$ and zeroes everywhere else. Then ${\sum_j E_j = I}$ and ${E_jE_l= 0}$ when ${j\neq l}$. Since ${A}$ has ${n}$ distinct real eigenvalues ${\lambda_1, \dots, \lambda_n}$, it is diagonalizable over ${\mathbf R}$, so there is a real matrix ${P}$ such that ${P^{-1}AP = D}$, where ${D=\mathrm{diag}(\lambda_1, \dots, \lambda_n) = \sum_j \lambda_j E_j }$. Let ${I_j = PE_jP^{-1}}$. Then

$\displaystyle \sum_j \lambda_j I_j = P\left(\sum_j \lambda_j E_j\right) P^{-1} = PDP^{-1} = A.$

Moreover, for ${j \neq l}$ we have ${I_jI_l = PE_jE_lP^{-1} = 0}$.

For the second part, notice that if the matrix ${A}$ is decomposed in the manner described above, the numbers ${\lambda_j}$ are necessarily eigenvalues of ${A}$. Indeed, multiplying the equality ${\sum I_j = I}$ by ${I_l}$ and using that ${I_lI_j = 0}$ when ${l \neq j}$, we find that ${I_l^2=I_l}$. Hence, let ${v \in \mathbf R^n}$ be any nonzero vector. Since ${\sum_j I_j v = v}$, at least one of the terms in the sum is nonzero, say ${I_l v \neq 0}$. Then

$\displaystyle AI_lv = \sum_j \lambda_j I_j I_lv = \lambda_l I_l^2v = \lambda_l I_lv,$

and therefore ${I_lv}$ is an eigenvector of ${A}$ with eigenvalue ${\lambda_l}$. Thus, it is impossible for the matrix ${A}$ to have such a decomposition if, say, it has no real eigenvalues, for example

$\displaystyle A=\left(\begin{array}{ll} 0 & -1 \\ 1 & 0 \end{array}\right).$