Prove if det(A) = 0 and det(A -kI)=0 then k=0.

Question

It seems to be, since $detA = 0$, If $det(A-kI)=0$ the kI part should be zero. So we can take k = 0 because I is nonzero. But I haven't any idea to take mathematical representation to prove it.

Can you help me. :-)

Answer 1

Let $A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$. Then $\det(A)=\det(A-kI)=0$ is equivalent to $ad=bc=(a-k)(d-k)$. This yields $k(k-(a+d))=0$. We see that either $k=0$, or $k=a+d$. In the latter case, $k$ need not be zero. So we obtain counter examples.

For example, take $$ A=\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}, $$ and $k=2$. Then $\det(A)=\det(A-2I)=0$.

Answer 2

Just look the follwing arguments: for clear concept of it.

(1) if $ lemda$ is a eigen value of a matrix $A$ then we get a nonzero vector $v$ (known as the eigen vector) such that
$A*v=lemda*v$
Then it can be rewritten as $ (A- lemda* indentity matrix)*v=0$ Looking carefully you can see that it is a homogeneous system of equation. For non trivial solution ( since the trivial solution would not work for us as we have nonzero vector $v$ )of it we should have

$ det (A-lemda * identity matrix)=0$

(2) Also you should know that the product of the eigen values of a matrix is equal to the determinent of that matirx.

Since det(A) is zero so one of the eigen values of the matrix $A$ is zero.But this does not tell that all the eigen values is zero. Which can be seen by the early users counterexample.

Answer 3

This isn't true. Take a look at $A:=\begin{pmatrix}1 & 1 \\ 1 & 1 \end{pmatrix}$. Then you can find solution for $\det{(A-kI)}=0$ for $k_{\pm} = 1 \pm 1$ ($k_+ = 2, k_-=0)$. You see there exists a solution with $k \neq 0$