1

$$ \begin{bmatrix} -2 & -2 & 0 & 0 \\ -b & -b & 0 & 0 \\ 0 & 0 & -4 & 4 \\ 0 & 0 & b & -b \\ \end{bmatrix} $$

he given matrix is a blocks one, so it is diagonalizable iff each block is, and the characteristic polynomials of each block are:

$$\begin{cases}\;\text{Upper Block:}&\;p_1(x)=(x+2)(x+b)-2b=x(x+b+2)\\{}\\ \text{Lower Block:}&p_2(x)=(x+4)(x+b)-4b=x(x+b+4)\end{cases}$$

Thus: in case $\;b=-2(\text{ or}\;-4)\;$, the upper block or the lower block isn't diagonalizable, as then we get a nilpotent non-zero matrix. In any other case ( i.e., $\;b\neq-2,-4\;$) , both blocks have two different eigenvalues and are thus diagonalizable, and then also the $\;4\times4\;$ matrix is.

then I need to find thee igenvector for eigenvalue 0 which $$V1=32$$ $$V2+V3=−12$$

MathIsHard
  • 11
  • Eigenvectors of zero are elements of the null space of the matrix. Computing null spaces is a basic skill that you need to to have in order to compute eigenvectors, and I would’ve thought that it had already been covered in whatever material it is that you’re studying. Review that, and then give this task another try on your own. – amd Jun 30 '19 at 00:18

0 Answers0