# If a matrix A's columns are linearly independent, are the columns of A-lambda*I also linearly independent?

Are there any cases such that given a matrix A with linearly independent columns, when subtracting from A some scalar multiple of the identity matrix, the resulting matrix's columns are linearly dependent?

And also, in the other case, if a matrix's columns are dependent and a scalar of the identity matrix is subtracted from it, could the result have linearly independent columns?

Thanks

• Any time $\lambda$ is an eigenvalue, this will be true Commented Mar 18, 2018 at 18:16

• Let $A$ be the identity matrix, let $\lambda =1$, then $A-\lambda I$ is the zero matrix, hence the columns are lienarly dependent.
• Let $A$ be the zero matrix (linearly dependent columns), let $\lambda =-1$, then $A-\lambda I$ is the identity matrix (linearly independent columns).
Yes, $A-\lambda I$ has linearly dependent columns iff $\lambda$ is an eigenvalue for $A$. So there is always such a $\lambda\in \Bbb C$ if your matrix has complex entries.