# Eigenvalue of an $n \times n$ real symmetric matrix with rank 2

Below is a question from the GATE Exam.

Let A be an $n \times n$ real valued square symmetric matrix of rank 2 with $\sum_{i=1}^{n} \sum_{j=1}^{n}A_{ij}^2=50$. Consider the following statements

(I) One Eigenvalue must be in [-5,5]

(II) The eigenvalue with the largest magnitude must be strictly greater than 5.

Which of the above statements about eigenvalues of A is/are necessarily CORRECT?

(A) Both I and II

(B) I only

(C) II only

(D) Neither I nor II

My attempt:

Let A be $\begin{bmatrix} -5&0\\0&5\\ \end{bmatrix}$ so, it's eigenvalues are -5 and 5. So statement I is true but II is false. So, the answer is B.

I can understand that the questions ask for conditions which are always true for a real values square symmetric matrix of rank 2.

Is there any better way to solve this?

• Your example only shows that II is not necessarily true. You cannot conclude that the answer is B, because you haven't eliminated D as a possible choice. – user1551 Jul 16 '18 at 5:29

## 3 Answers

We are being told that $\|A\|_F^2 =50$ and also $A$ is a real symmetric matrix with rank $2$.

Hence $\sum_{i=1}^n \lambda_i^2 =50$ and we know that the eigenvalues are real.

Suppose all eigenvalues are not in $[-5, 5]$, that is if $|\lambda_i|>5, \forall i \in \{1,\ldots, n\}$ . Then, $$\sum_{i=1}^n \lambda_i^2 \ge n \min_i \lambda_i^2> 2(5)^2=50.$$

which is a contradiction. Hence $(I)$ is true.

• -What is this $||A||_F^2$ ? – user3767495 Jul 16 '18 at 9:16
• Frobenius norm, basically sum of squares of each element. – Siong Thye Goh Jul 16 '18 at 9:23

Take the matrix$A$ of the form $A= \operatorname{diag} ( 5,5,0,... )$. Then $\|A\|_F^2 = 50$, but both eigenvalues are in $[-5,5]$. Hence II is not true.

Since $A$ is symmetric, the singular values are the absolute values of the eigenvalues (which are real). Since $A$ has rank 2 there are exactly two non zero eigenvalues.

Hence $\|A\|_F^2 = \lambda_1^1+\lambda_2^2 = 50$. Hence $\min(\lambda_1^1,\lambda_2^2) \le 25$ and so $\min(|\lambda_1|,|\lambda_2|) \le 5$ and so $I$ is true.

Hence B is the correct answer.

If $A = \begin{bmatrix} -5 & 0 \\ 0 & 5 \end{bmatrix}$ then $\|A\|_{F}^{2}\neq 50.$ If $A$ is a diagonal matrix then $\|D\| = \max_{1 \leq i \leq n } |d_{i}|$ so $\|A \|_{F}^{2} = 5$

A = [-5,0;0,5];
my = norm(A);
display(my)
my =

5


Also, the eigenvalues of a triangular matrix are the diagonal. And diagnonal matrices triangular matrices. So the eigenvalues are $-5,5$