Question

Let the eigen values of a 2 x 2 matrix A be 1, -2 with eigen vectors $X_1$ and $X_2$ respectively. Then the eigen values and eigen vectors of the matrix $A^2$ - 3A + 4I would, respectively, be

• A. 2, 14, $X_1,X_2$
• B. 2, 14, $X_1+X_2,X_1,-X_2$
• C. 2, 0, $X_1,X_2$
• D. 2, 0, $X_1+X_2,X_1-X_2$

Answer 1

The correct option is A 2, 14, $X_1,X_2$

From the property of eigen values, if ${\lambda }_1,{\lambda }_2,.........{\lambda }_n$ are the eigen values of a matrix A, then $A^m$ has the eigen values ${\lambda }_1^m,{\lambda }_2^m,......{\lambda }_n^m$ (m being a positive integer) and the corresponding eigen vector is the same. Hence, by using this property, eigen values of any polynomial of A can be obtained by replacing A by $\lambda$.

Hence the eigen values of the matrix $A^2$ - 3A + 4I will be

$(1{)}^2$ - 3(1) + 4 = 2 & $(-2{)}^2$ - 3(-2) + 4 = 14