Eigenvalues of matrix with linear dependent rows

Question

Why is it that, if a matrix has a linear dependent row, e.g.: $$ A=\begin {pmatrix} 1 \ 2 \\ 2 \ 4 \end {pmatrix}, $$ at least one eigenvalue is zero?

Answer 1

Ok, so I found an answer with the help of Bye_World's comment and want to provide it in case anybody else is also wondering: In order to find eigenvalues of a matrix, we need its characteristic polynomial, which is calculated as follows: $$ \det (\lambda I-A)=\det \begin {pmatrix} \lambda -a_{11} & -a_{12} \\ -a_{21} & \lambda -a_{22} \end{pmatrix}=\lambda ^2 - (a_{11}+a_{22}) \lambda + a_{11}a_{22}-a_{12}a_{21}=0, $$ where $$ \det (A)= a_{11}a_{22}-a_{12}a_{21}. $$ So we can see that if $det (A)=0$, which is the case if there is a linear dependent row, then $\lambda=0$ is a solution.

Answer 2

If λ=0, then A=A-λI, and det(A)=det(A-λI).

There's a linear dependent row (as you said) in A, and that implies that det(A)=0. But det(A)=det(A-λI), so det(A-λI)=0, and λ=0 is an eigenvalue of A.