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
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.
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
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
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
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
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 = \(\begin{bmatrix}
0 & -1 & 0 \\
1 & 0 & -1\\
0 & 1 & 1
\end{bmatrix} \)
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
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
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
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
Explanation: The product of the eigen values of a matrix gives the determinant of the matrix,
Therefore, ∆ = 24.
10. Which of the following relation is correct?
a) A = AT
b) A = -AT
c) A = A2T
d) A = AT/2
View Answer
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.
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