I’ve encountered the following question, which I’m having trouble trying to solve:
Given a non-invertible matrix $A$, prove it is similar to:
- A matrix with a zero column
- A matrix with a zero row
I have no idea how to prove the first part. As to the second part, I know the row echelon form of $A$ has a zero row so $A=EP$ where $P$ is the ref and $E$ is an invertible matrix. However this is not the definition of similarity.
Any help will be appreciated, thanks in advance!
Just to clarify, I think you are talking about square matrices here.
Suppose the columns of the non-invertible matrix are $c_1, \cdots, c_n$. Then these vectors are not linearly independent and hence there exists scalars $a_1, \cdots, a_n$, not all zero, such that $a_1c_1 + \cdots +a_n c_n = 0$, the zero vector. Suppose $a_i \ne 0$, then divide the equation by $a_i$ to get $\frac{a_1}{a_i}c_1 + \cdots + c_i + \cdots + \frac{a_n}{a_i}c_n = 0$. Now consider the following transformation, $c_i \leftrightarrow \frac{a_1}{a_i}c_1 + \cdots + c_i + \cdots + \frac{a_n}{a_i}c_n$, which is adding a linear combination of the other vectors to the $i$-th column of the matrix, which is equivalent to postmultiplying the matrix with an invertible matrix, say $E$. So we have $AE = B$, where $B$ has all the columns same as $A$ except the $i$-th column which is $\frac{a_1}{a_i}c_1 + \cdots + c_i + \cdots + \frac{a_n}{a_i}c_n$, which is nothing but the zero vector. Now premultiplying $B$ by any matrix will leave the $i$-th column unchanged, equal to the zero vector. In particular we premultiply by $E^{-1}$, say $E^{-1}B = C$, where $C$ also has its $i$-th column equal to the zero vector. Hence $C = E^{-1}A E$ is a matrix similar to $A$, that has a zero column.
The same argument holds for the rows, which also cannot be linearly independent.