Home » Linear Algebra » Ph.D. Comprehensive exam practice problems, Round 1

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

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

Problem: Let ${f \geq 0}$ be a continuous function on ${[0,1]}$. Suppose there exists ${C \in \mathbf R}$ such that for all ${x \in [0,1]}$,

$\displaystyle f(x) \leq C \int_0^x f(t) dt.$

Prove that ${f=0}$.

Solution: (With Juan Ignacio Restrepo.) By iterated integration, we find by induction that, for every integer ${n \geq 0}$,

$\displaystyle f(x) \leq \frac{C^{n+1}}{n!} \int_0^x (x-t)^n f(t) dt.$

Since the integrand is uniformly bounded and ${C^{n+1}/n! \rightarrow 0}$ as ${n \rightarrow \infty}$, it follows that ${f = 0}$.

Problem: Suppose that ${\{a_n\}}$ is a real sequence with ${a_n \rightarrow 0}$ and that ${\{b_n\}}$ is a nonnegative sequence with ${\lim_{n\rightarrow \infty} (b_1 + \dots + b_n) = \infty}$. Prove that

$\displaystyle \lim_{n \rightarrow \infty} \left(\frac{a_1b_1 + \dots + a_nb_n}{b_1+ \dots + b_n}\right) = 0.$

Solution: For notational simplicity, let us write ${s_n = b_1 + \dots + b_n}$. Let ${N_1}$ be so large that ${|a_n|< \epsilon/2}$ for all ${n>N_1}$. Then, for ${n>N_1}$, we have, by the triangle inequality,

$\displaystyle \left|\frac{1}{s_n} (a_1b_1 + \dots + a_nb_n) \right|\leq \frac{1}{s_n}(C + \frac{\epsilon}{2}(s_n - s_{N_1})) = \frac{C-s_{N_1}}{s_n} + \epsilon/2,$

where ${C = |a_1b_1 + \dots + a_{N_1}b_{N_1}|}$. Let ${N_2}$ be so large that ${|C-s_{N_1}|<\epsilon s_n / 2}$ for all ${n>N_2}$, which exists since ${s_n \rightarrow \infty}$. Then for ${n> \max({N_1, N_2})}$, we have

$\displaystyle \left|\frac{1}{s_n} (a_1b_1 + \dots + a_nb_n)\right| < \epsilon.$