Linear Algebra Questions and Answers – Diagonalization Powers of a Matrix

«
»

This set of Linear Algebra and Vector Calculus Multiple Choice Questions & Answers (MCQs) focuses on “Diagonalization Powers of a Matrix”.

1. Which of the following is not a necessary condition for a matrix, say A, to be diagonalizable?
a) A must have n linearly independent eigen vectors
b) All the eigen values of A must be distinct
c) A can be an idempotent matrix
d) A must have n linearly dependent eigen vectors
View Answer

Answer: d
Explanation: The theorem of diagonalization states that, ‘An n×n matrix A is diagonalizable, if and only if, A has n linearly independent eigenvectors.’ Therefore, if A has n distinct eigen values, say λ1, λ2, λ3…λn, then the corresponding eigen vectors are said to be linearly independent. Also, all idempotent matrices are said to be diagonalizable.
advertisement

2. The geometric multiplicity of λ is its multiplicity as a root of the characteristic polynomial of A, where λ be the eigen value of A.
a) True
b) False
View Answer

Answer: b
Explanation: The diagonalization theorem in terms of multiplicities of eigen values is defined as follows,
The algebraic multiplicity of λ is its multiplicity as a root of the characteristic polynomial of A.
The geometric multiplicity of λ is the dimension of the λ-eigenspace.

3. If A is diagonalizable then, ____________
a) An = (PDP-1)n = PDnPn
b) An = (PDP-1)n = PDnP1
c) An = (PDP-1)n = PDnP-1
d) An = (PDP-1)n = PDnP
View Answer

Answer: c
Explanation: The definition of diagonalization states that, An n × n matrix A is diagonalizable if there exists an n × n invertible matrix P and an n × n diagonal matrix D such that,
P-1 AP = D
A = PDP-1
An = (PDP-1)n = PDnP-1

4. The computation of power of a matrix becomes faster if it is diagonalizable.
a) True
b) False
View Answer

Answer: a
Explanation: Some of the applications of diagonalization of a matrix are:
The powers of a diagonalized matrix can be computed easily since the result is nothing but the powers of the diagonal elements obtained by diagonalization.
Reducing quadratic forms to canonical forms by orthogonal transformations.
In mechanics, it can be used to find the natural frequency of vibrations.

5. Find the invertible matrix P, by using diagonalization method for the following matrix.
A = \(\begin{bmatrix}
2 & 0 & 0 \\
1 & 2 & 1\\
-1 & 0 & 1
\end{bmatrix} \)
a) A = \(\begin{bmatrix}
-1 & -1 & 0 \\
1 & 0 & -1\\
-1 & 1 & 1
\end{bmatrix} \)
b) A = \(\begin{bmatrix}
0 & -1 & 0 \\
1 & 0 & -1 \\
0 & 1 & 1
\end{bmatrix} \)
c) A = \(\begin{bmatrix}
0 & 0 & 0\\
1 & 1 & -1\\
-1 & 0 & 1
\end{bmatrix} \)
d) A = \(\begin{bmatrix}
1 & 0 & 0\\
1 & 0 & -1\\
-1 & 0 & 1
\end{bmatrix} \)
View Answer

Answer: b
Explanation: Procedure to find the invertible matrix is as follows,
Step 1: Find the eigen values of the given matrix.
A = \(\begin{bmatrix}
2 & 0 & 0 \\
1 & 2 & 1\\
-1 & 0 & 1
\end{bmatrix} \)
⎸A – λI ⎸ = 0
\(\begin{vmatrix}
2-λ & 0 & 0\\
1 & 2-λ & 1\\
-1 & 0 & 1-λ
\end{vmatrix} \) = 0 ……………………… (i)
(2-λ) ((2- λ) (1- λ)) = 0
(2- λ)2 (1-λ) = 0
λ = 2, 2, 1
Step 2: Compute the eigen vectors
Consider λ = 2,
(A – λI) \(\vec{X} = \vec{0} \)
\(\begin{bmatrix}
0 & 0 & 0\\
1 & 0 & 1 \\
-1 & 0 & -1
\end{bmatrix}
\vec{X} = \vec{0} \quad ^{\underrightarrow{Reducing \,further, \,we get,\,}}
\begin{bmatrix}0 & 0 & 0 \\
1 & 0 & 1 \\
0 & 0 & 0\end{bmatrix}
\vec{X}=\vec{0}
\)
x2 is the free variable, hence, x2 = s
Let x1 = -t, x3 = t, since x1+x3 = 0
\(\vec{X} = s\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix} \) + t\(\begin{bmatrix}
-1\\
0\\
1
\end{bmatrix} \)
\(\vec{X_1} = \begin{bmatrix}
0\\
1\\
0
\end{bmatrix} \) \(\vec{X_2} = \begin{bmatrix}
-1\\
0\\
1\end{bmatrix} \)
Consider λ = 1
(A – λI)\(\vec{X} = \vec{0} \)
\(\begin{bmatrix}
1 & 0 & 0\\
1 & 1 & 1 \\
-1 & 0 & 0
\end{bmatrix}
\vec{X} = \vec{0} \quad ^{\underrightarrow{Reducing \,further, \,we get,\,}}
\begin{bmatrix}1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 0 & 0\end{bmatrix}
\vec{X}=\vec{0}
\)
x1 = 0, Let x2 = -s and x3 = s since x2+x3=0
\(\vec{X}= s\begin{bmatrix}
0\\
-1\\
1
\end{bmatrix} \)
\(\vec{X_3} = \begin{bmatrix}
0\\
-1\\
1
\end{bmatrix} \)
Step 3: Formation of the invertible matrix.
\(P = [\vec{X_1} \vec{X_2} \vec{X_3}]\)
P = \(\begin{bmatrix}
0 & -1 & 0 \\
1 & 0 & -1\\
0 & 1 & 1
\end{bmatrix} \)
advertisement

6. Determine the algebraic and geometric multiplicity of the following matrix.
\(\begin{bmatrix}
2 & 4 & -4 \\
0 & 4 & 2\\
-2 & 4 & 4
\end{bmatrix} \)
a) Algebraic multiplicity = 1, Geometric multiplicity = 2
b) Algebraic multiplicity = 1, Geometric multiplicity = 3
c) Algebraic multiplicity = 2, Geometric multiplicity = 2
d) Algebraic multiplicity = 2, Geometric multiplicity = 1
View Answer

Answer: d
Explanation: The eigen values of the given matrix can be computed as,
⎸A – λI ⎸ = 0
\(\begin{vmatrix}
1-λ & 0 & 1\\
3 & 3-λ & 0\\
0 & 0 & 1-λ
\end{vmatrix}\)= 0
(1-λ) ((3-λ) (1-λ)) = 0
(1-λ)2 (3-λ) = 0
λ = 1, 1, 3 are the eigen values of the matrix. So, the algebraic multiplicity of λ = 1 is two.
For λ = 3,
(A – λI) \(\vec{X} = \vec{0} \)
\(\begin{bmatrix}
-2 & 0 & 1 \\
3 & 0 & 0 \\
0 & 0 & -2
\end{bmatrix}\vec{X} = \vec{0}\quad ^{\underrightarrow{Reducing \,further, \,we \,get,}}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & 1\\
0 & 0 & 0
\end{bmatrix}
\vec{X} = \vec{0} \)
x1 = 0, x3 = 0
x2 is the free variable, therefore let x2 = s,
Hence, \(\vec{X_1}= s\begin{bmatrix}0 \\ 1\\ 0\end{bmatrix} ≈ \begin{bmatrix}0 \\ 1 \\0 \end{bmatrix} \)
For λ = 1,
(A – λI) \(\vec{X} = \vec{0} \)
\(\begin{bmatrix}
0 & 0 & 1 \\
3 & 2 & 0 \\
0 & 0 & 0
\end{bmatrix} \vec{X} = \vec{0} \quad ^{\underrightarrow{Reducing \,further, \,we \,get,}}
\begin{bmatrix}
1 & 2/3 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{bmatrix} \vec{X} = \vec{0} \)
x1 + \(\frac{2}{3}\) x2 = 0, x3 = 0
Let x1 = -2 and x3 = 3,
\(\vec{X_2} = \begin{bmatrix}-2 \\ 3\\ 0 \end{bmatrix}\)
Thus, there corresponds only one eigen vector for the repeated eigen value λ=1. Thus, the geometric multiplicity of λ = 2 is one.

7. Given P = \( \begin{bmatrix}2 & -1 \\ 5 & 1 \end{bmatrix} \, and\, D = \begin{bmatrix}6 & 0 \\ 0 & -1\end{bmatrix},\) find A3.
a) \(\begin{bmatrix}
61 & 62 \\ 156 & 154\end{bmatrix}\)
b) \(\begin{bmatrix}
61 & 62 \\ 155 & 154 \end{bmatrix}\)
c) \(\begin{bmatrix}
61 & 60 \\ 155 & 154 \end{bmatrix}\)
d) \(\begin{bmatrix}
61 & 62\\ 155 & 150\end{bmatrix}\)
View Answer

Answer: b
Explanation: From the theory of diagonalization, we know that,
A = PDP-1
An = PDnP-1
Given, P= \(\begin{bmatrix}
2 & -1\\
5 & 1,\end{bmatrix} hence P^{-1} = \frac{1}{7}
\begin{bmatrix}
1 & 1\\
-5 & 2\end{bmatrix}\)
Therefore, \(A^3 = \frac{1}{7} \begin{bmatrix}
2 & -1\\
5 & 1
\end{bmatrix}
\begin{bmatrix}
6 & 0\\
0 & -1 \end{bmatrix}^3
\begin{bmatrix}
1 & 1\\
-5 & 2 \end{bmatrix}\)…………………………. since n=3
\(A^3 = \frac{1}{7}
\begin{bmatrix}2 & -1 \\ 5 & 1 \end{bmatrix}
\begin{bmatrix}216 & 0 \\ 0 & -1 \end{bmatrix}
\begin{bmatrix}1 & 1 \\ -5 & 2 \end{bmatrix}\)
\(A^3 = \frac{1}{7}
\begin{bmatrix} 2 & -1 \\ 5 & 1 \end{bmatrix}
\begin{bmatrix} 216 & 216 \\ 5 & -2 \end{bmatrix}\)
\(A^3 = \frac{1}{7}
\begin{bmatrix}427 & 434\\ 1085 & 1078 \end{bmatrix}\)
\(A^3 = \begin{bmatrix} 61 & 62\\ 155 & 154 \end{bmatrix}\)

8. Find the trace of the matrix \(A = \begin{bmatrix}1 & 0 & 6\\
0 & 5 & 0\\
0 & 4 & 4 \end{bmatrix}.\)
a) 0
b) 10
c) 4
d) 1
View Answer

Answer: b
Explanation: The sum of the entries on the main diagonal is called the trace of matrix A.
Therefore, trace = 1+5+4 = 9.

9. The determinant of the matrix whose eigen values are 4, 2, 3 is given by, _______
a) 9
b) 24
c) 5
d) 3
View Answer

Answer: b
Explanation: The product of the eigen values of a matrix gives the determinant of the matrix,
Therefore, ∆ = 24.
advertisement

10. Which of the following relation is correct?
a) A = AT
b) A = -AT
c) A = A2T
d) A = AT/2
View Answer

Answer: a
Explanation: To prove that A = AT, let us consider an example,
\(A = \begin{bmatrix} 1 & 0 \\ 2 & 3 \end{bmatrix}\)
⎸A – λI ⎸ = 0
\( \begin{vmatrix} 1-λ & 0 \\ 2 & 3-λ \end{vmatrix} = 0\)
(1-λ) (3-λ) = 0
3 – λ – 3λ +λ2 = 0
λ2 – 4λ + 3 = 0
(λ – 3) (λ – 1) = 0
λ = 1, 3
Consider \(A^T = \begin{bmatrix}1 & 2\\ 0 & 3 \end{bmatrix}\)
⎸A – λI ⎸ = 0
\(\begin{bmatrix}1-λ & 2 \\ 0 & 3-λ \end{bmatrix} = 0\)
(1-λ) (3-λ) = 0, which is similar to the result obtained for A, hence the eigen values are same.

Sanfoundry Global Education & Learning Series – Linear Algebra and Vector Calculus.

To practice all areas of Linear Algebra and Vector Calculus, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
advertisement

Leave a Comment

Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn