CSIR NET JUNE 2020 LINEAR ALGEBRA [Under Progress]

Problem 1

Let \( A \) be a \( 2 \times 2 \) real matrix with \( \det A = 1 \) and \( \operatorname{tr} A = 3 \). What is the value of \( \operatorname{tr} A^{2} \)?

Options:
(1) 2
(2) 10
(3) 9
(4) 7

Correct Answer: (4)

📖 View Step-by-Step Solution

Method 1: Eigenvalue Approach

Let \( \lambda_1 \) and \( \lambda_2 \) be eigenvalues of \( A \):

\[ \begin{aligned} \operatorname{tr}(A)= \lambda_1 + \lambda_2 &= 3 \\ \operatorname{det}(A)= \lambda_1 \lambda_2 &= 1 \end{aligned} \]

Calculate \( \operatorname{tr}(A^2) \):

\[ \begin{aligned} \operatorname{tr}(A^2) &= \lambda_1^2 + \lambda_2^2 \\ &= (\lambda_1 + \lambda_2)^2 - 2\lambda_1\lambda_2 \\ &= 3^2 - 2(1) \\ &= 9 - 2 = 7 \end{aligned} \]

Method 2: Characteristic Polynomial

Using the characteristic equation:

\[ x^2 - (\operatorname{tr}A)x + \det A = 0 \\ \Rightarrow x^2 - 3x + 1 = 0 \]

Solving for eigenvalues:

\[ \lambda = \frac{3 \pm \sqrt{5}}{2} \]

Calculate squared eigenvalues:

\[ \lambda_1^2 + \lambda_2^2 = \left(\frac{3+\sqrt{5}}{2}\right)^2 + \left(\frac{3-\sqrt{5}}{2}\right)^2 = 7 \]

Problem 2

For any two \( 2 \times 2 \) matrices \( A \) and \( B \), which statement is always true?

Options:
(1) \(\operatorname{det}(A+B) + \operatorname{det}(A-B) = \operatorname{det}A + \operatorname{det}B\)
(2) \(\operatorname{det}(A+B) + \operatorname{det}(A-B) = 2\operatorname{det}A - 2\operatorname{det}B\)
(3) \(\operatorname{det}(A+B) + \operatorname{det}(A-B) = 2\operatorname{det}A + 2\operatorname{det}B\)
(4) \(\operatorname{det}(A+B) - \operatorname{det}(A-B) = 2\operatorname{det}A - 2\operatorname{det}B\)

Correct Answer: (3)

📖 Show Solution Methods

Method 1: Algebraic Expansion

Let \( A = \begin{pmatrix}a&b\\c&d\end{pmatrix} \) and \( B = \begin{pmatrix}w&x\\y&z\end{pmatrix} \):

Step 1: Calculate \(\operatorname{det}(A+B)\):
\[ (a+w)(d+z) - (b+x)(c+y) \]

Step 2: Calculate \(\operatorname{det}(A-B)\):
\[ (a-w)(d-z) - (b-x)(c-y) \]

Step 3: Add both determinants:
\[ \begin{align*} = & [ad + az + wd + wz - bc - by - xc - xy] \\ + & [ad - az - wd + wz - bc + by + xc - xy] \\ = & 2ad - 2bc + 2wz - 2xy \\ = & 2(\operatorname{det}A + \operatorname{det}B) \end{align*} \]

Method 2: Example Verification

Test Case 1:
Let \( A = I = \begin{pmatrix}1&0\\0&1\end{pmatrix} \), \( B = 0 \)
\(\operatorname{det}(A+B) + \operatorname{det}(A-B) = 1 + 1 = 2\)
\(2(\operatorname{det}A + \operatorname{det}B) = 2(1 + 0) = 2\) ✔️

Test Case 2:
Let \( A = I \), \( B = -I \)
\(\operatorname{det}(A+B) + \operatorname{det}(A-B) = 0 + 4 = 4\)
\(2(\operatorname{det}A + \operatorname{det}B) = 2(1 + 1) = 4\) ✔️

Problem 3

Which quadratic form on \(\mathbb{R}^2\) is positive definite?

Options:
(1) \(Q(x, y) = xy\)
(2) \(Q(x, y) = x^2 - xy + y^2\)
(3) \(Q(x, y) = x^2 + 2xy + y^2\)
(4) \(Q(x, y) = x^2 + xy\)

Correct Answer: (2)

📘 Show Solution Methods

Recall:
1. A real qaudratic form \(Q\) on \(R^{n}\) is said to positive definite if \(Q(x) > 0\), for \(x\neq 0\).
1. A real symmetric matrix is positive defininite IF AND ONLY IF its all eigen values >0

Method 1: Direct Evaluation

Option (1): \(Q(1,-1) = (1)(-1) = -1\) (Negative ⇒ Not positive definite)

Option (3): \(Q(1,-1) = 1 - 2 + 1 = 0\) (Zero ⇒ Not positive definite)

Option (4): \(Q(1,-1) = 1 - 1 = 0\) (Zero ⇒ Not positive definite)

Option (2): \(Q(x,y) = x^2 - xy + y^2\)
Completing the square: \(= \left(x - \frac{y}{2}\right)^2 + \frac{3y^2}{4}\)
Always positive for \((x,y) \neq (0,0)\)

Method 2: Matrix Approach

Option	Matrix	Conclusion on Eigenvalues
(1)	\(\begin{bmatrix}0 & 0.5 \\ 0.5 & 0\end{bmatrix}\)	Determinant negative ⇒ one eigen value negative ⇒ Not positive definite
(2)	\(\begin{bmatrix}1 & -0.5 \\ -0.5 & 1\end{bmatrix}\)	\(\frac{3}{2}, \frac{1}{2}\) ⇒ Positive definite
(3)	\(\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}\)	Determinant=0 ⇒ One eigen value =0 ⇒ Not positive definite
(4)	\(\begin{bmatrix}1 & 0.5 \\ 0.5 & 0\end{bmatrix}\)	Determinant negative ⇒ one eigen value negative ⇒ Not positive definite

Problem 4

Let \( A \) be an \( n \times n \) matrix with exactly \( r \) distinct non-zero eigenvalues. Which statement is always true?

Options:

1. \(\operatorname{rank}(A) \leq r\)
2. If \( r=0 \), then \( \operatorname{rank} A < n-1 \)
3. \(\operatorname{rank}(A) \geq r\)
4. \( A^{2} \) has \( r \) distinct non-zero eigenvalues

Correct Answer: (3)

📚 Detailed Solution

Method 1: Counterexample Analysis

Option (2) and option (1) Analysis:

Consider \( A = \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix} \):

• All eigenvalues = 0 ⇒ \( r = 0 \)
• But \( \operatorname{rank}(A) = 1 = 2-1 \)

This violates option (2), making it false.

Also since \( \operatorname{rank}(A) = 1\) and \(r=0\), so option (1) also false.

Option (4) Analysis:

For \( A = \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix} \):

• Eigenvalues: \( \pm i \) (non-real)
• \( A^2 \) has eigenvalues: -1, -1 (repeated)

Original \( r=2 \), but \( A^2 \) has only 1 distinct eigenvalue. Hence option (4) fails.

Method 2: Jordan Form Approach

Any matrix can be transformed to Jordan Canonical Form (which is an upper traingular matrix):

\[ J = \begin{bmatrix} \lambda_1 & * & \dots & \dots & \dots & * \\ & \ddots & & \\ & & \lambda_r & * & \dots &* \\ & & & 0 & \dots & * \\ & & & & \ddots \\ & & & & & 0 \end{bmatrix} \]

observations:

• Hence the matrix \(J\) has atleast \(r\) linearly independent rows.
• Hence \( \operatorname{rank}(A) =\operatorname{rank}(J)\geq r \).
• So option (3) is correct.