GO Classes CS/DA 2025 | Weekly Quiz 4 | Linear Algebra | Question: 30

Answer 1

Properties of a Real Symmetric matrix.

1. $N\times N$ Real symmetric matrices have $N$ Real eigenvalues.
2. Eigenvectors of distinct eigenvalues are orthogonal.

$Statement$ 1: Here, Eigenvectors for $S$ are $\begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}$ with Eigenvalue 2 and $\begin{bmatrix} 2\\ 1\\ 0 \end{bmatrix}$ with eigenvalue -1. In this case the eigenvalues are distinct but the dot product of these Eigenvectors is not zero. 1*2 + 2*1 + 3*0 = 4. Therefore, these two eigenvectors are not orthogonal and hence statement 1 is false.

$Statement$ 2:  Given, $S^{2} + I = 0$. Applying the Cayley-Hamilton theorem, we observe that $S$ satisfies the underlying charecterisitc equation $\lambda ^{2} = -1$. Here Eigenvalues of $S$ are $i$ and $-i$ which are complex numbers. Hence, $Statement$ 2 is false.

Comments

Eighen value of S is -1 so Eighen value of S^2 is 1.S^2 should satisfy its charecteristic equation S^2 =I. Is this Right?

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@Shreyash007 we cannot say anything about the characteristic equation of S in statement 1. Because we need the third eigenvalue which is not given. If the third eigenvalue is a then the characteristic equation of S in statement 1 would be ($\lambda$ – a)($\lambda$ +1)($\lambda$ – 2) = 0 and S satisfies this equation.

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how eigen values are i and -i for st 2

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@AGNIDEB MUKHERJEE

Solve the quadratic equation $\lambda ^{2} + 1 = 0$ and find values of $\lambda$.

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Ok got it thanks 👍

Answer 2

Statement 1 has two eigenvectors with different eigenvalues.

Since it is a symmetric matrix hence both eigenvectors corresponding to  different eigenvalues must be orthogonal.

 

For Statement 2, can you take determinant both side and check if it is possible ?
Anyone want to try ?

Comments

For stmt 2 we don’t know what is S then how to do S^2 even

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@Omi Kakadiya S^2 = -I, now taking determinant on both sides, |S^2| = |-I| → |S|^2 = -1. How can the square of the determinant of a matrix with real values be -1? It is not possible. Hence that statement is false.