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.

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