369 Midterm 2
Your Name:
Instructions: This test contains 5 questions. Each question is worth 20 points. Read each
question carefully before responding. Give justifications to all of your responses, unless
otherwise specified. No calculators are allowed. One standard sized page of notes with
writing on one side is allowed.
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1. Find the eigenvectors and eigenvalues of


2 0 0
A = 0 1 1 .
0 1 1
Solution. The eigenvalues are λ = 0, 2, and the eigenvectors are
     
0
0
1
−1 , 1 , 0 .
1
1
0
2
2. Let
A=
1 2
.
0 3
Find an invertible matrix P and a diagonal matrix C such that A = P CP −1 .
Solution. By finding eigenvectors and eigenvalues of A, we get
1 1
1 0
P =
,C=
.
0 1
0 3
3
3. Consider the matrix


2 2 0
A = 0 4 0  .
0 −3 4
(i) Show that 4 is an eigenvalue of A.
Solution. Note that pA (λ) = (λ − 2)(λ − 4)2 = 0 when λ = 4.
(ii) Find the eigenspace corresponding to this eigenvalue.
 
 0 
Solution. The eigenspace is the kernel of A − 4I, which is span 0 .


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(iii) Find the minimal polynomial qA (λ) of A.
From part (ii) and the fact that geometric multiplicity ≤ algebraic multiplicity, we know that the
geometric multiplicity of both eigenvalues is 1. Since 1+1 < 3, we know A is not diagonalizable.
By a theorem, qA cannot be a product of distinct linear factors. Since qA divides pA , we can
conclude that qA (λ) = (λ − 2)(λ − 4)2 .
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4. Suppose the characteristic polynomial pA (λ) and the minimal polynomial qA (λ) of A satisfy
pA (λ) = (λ − 1)(λ + 2)2 = qA (λ).
(i) What is the size of the matrix A?
Solution. It is 3 × 3, since the degree of pA is 3.
(ii) What are the eigenvalues of A?
Solution. 1 and −2, the roots of pA .
(iii) What is the algebraic multiplicity of each eigenvalue?
Solution. The algebraic multiplicities of 1 and −2 can be read off from pA : they are 1 and 2,
respectively.
(iv) What is the geometric multiplicity of each eigenvalue?
Solution. Since qA is not a product of distinct linear factors, we know that A is not diagonalizable and so the sum of the geometric multiplicities of the eigenvalues is < 3. Thus, the only
possibility is that both geometric multiplicities equal 1.
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5. Let A be a square matrix. Prove that Ker A ⊂ Ker A2 .
Solution. Suppose x ∈ Ker A. Then Ax = 0, so A2 x = A(Ax) = A0 = 0, so x ∈ Ker A2 .
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