Is this a diagonalizable matrix?

Question

question:

Suppose $A$ is a $3\times3$ matrix such that $\det(A)=0$, $\det(A+2I)=0$, $\det(A-3I)=0$.

1. Is $A$ diagonalizable?
2. Is $A$ invertible?
3. What is the rank of $A$?

1. So the solution tells me that $A$ is diagonalizable since it has $3$ distinct eigenvalues: $0$, $-1$, and $3$. But i don't understand how you get those $3$ eigenvalues. I thought that the eigenvalues would be $3$ and $-2$ since $-2$ satisfies $\det(A+2I)$ and $3$ satisfies $\det(A-3I)$. I'm really confused how you can get those $3$ eigenvalues without doing the computation on a matrix.

2. I know that it's not invertible since it's provided that $\det(A) = 0$.

3. not sure how to really figure this one out either without the matrix.

If a matrix $A$ has $\det(A) = 0$, does that mean $0$ is one of it's eigenvalues?

Answer 1

The characteristic polynomial is $$\chi_A(x)=\det(A-xI)$$ and its roots are the eigenvalues of the matrix $A$. By the hypothesis we see that $0,-2$ and $3$ are the $3$ distinct eigenvalues so $A$ is diagonalizable and since $A$ has $0$ as eigenvalue with multiplicity $1$ then $A$ isn't invertible and $$\dim\ker A=1$$ so by the rank-nullity theorem the rank of $A$ is $2$.

Answer 2

By definition, $\lambda$ is an eigenvalue if there exists some nonzero $x$ such that $Ax=\lambda x$.

This is equivalent to the existance of a nonzero $x$ such that $(A-\lambda I) x = 0$, which is equivalent to $\det(A-\lambda I) = 0$.

Conclusion: if $\det(A-\lambda I) = 0$, then $\lambda$ is an eigenvalue of $A$.

Plugging $\lambda = 0$ into the conclusion also answers your final question.

Answer 3

The three determinants that are zero tell you that $0$, $-2$ and $+3$ are roots of the characteristic polynomial. Since your matrix is $3\times3$, the degree of that chacteristic polynomial is$~3$, which forces all those roots to be simple roots, and the characteristic polynomial to be $X(X+2)(X-3)$. In the case of simple roots (and only in that case) you can know without further computation what the dimension of the corresponding eigenspace will be, namely$~1$, so here you have eigenspaces for $0$, $-2$ and $+3$, each of dimension$~1$. Together they span the whole $3$-dimensional space, so $A$ is diagonalizable. The $1$-dimensional eigenspace for $\lambda=0$ is the kernel of $A$, which matrix is therefore is not invertible and has rank$~3-1=2$.

Note that all conclusions depend on having as much distinct eigenvalues as the dimension of the space; without that (for instance if only two of those zero determinants were given) nothing could have been said.