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I am trying to set up some "mental models" for how to think about matrix invertability. I am currently studying linear algebra on a basic level and I would please like some explanations to the question below, general information about matrix invertability related to this question is also much appreciated!

So if we have a matrix $A = \left( \begin{matrix} row_1 \\ row_2 \\ row_3 \end{matrix}\right )$

We are trying to find the matrix $A^{-1}$ to get to the matrix $I$.

When trying to find $I$ we get four cases where invertability breaks down:

1) So if the $row_1$ of matrix A is all zeros then it is obvious that we cannot find a matrix $A^{-1}$ that satisfies the following relationship: $\left( \begin{matrix} row_1 \\ row_2 \\ row_3 \end{matrix}\right ) A^{-1} \neq I$ since it will be impossible to produce the first column of $I$ using matrix multiplication.

2) The same is true if $col_1$ of $A$ is all zeros: $A^{-1} \left( \begin{matrix} col_1 & col_2 & col_3 \end{matrix}\right ) \neq I$ since it will be impossible to produce the first column of $I$ using matrix multiplication.

3) Now what I have a harder time to explain to myself is if we have $row_1$ of $A$ to be all zeros: $A^{-1} \left( \begin{matrix} row_1 \\ row_2 \\ row_3 \end{matrix}\right ) \neq I$ since it will be ...??

4) I also lack a clear explanation to the case where $col_1$ of $A$ is all zeros:

$\left( \begin{matrix} col_1 & col_2 & col_3 \end{matrix}\right ) A^{-1} \neq I$ since it will be ...??

Thank you for your time and help!

Lukas Arvidsson
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1 Answers1

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There is much more to a matrix being non-invertible than having a zero row or column. In order for some matrix $B$ to exist so that $BA=I$, every row of $I$ (and therefore every possible row) should be realisable as a linear combination of the rows of $A$. This will certainly not be the case if $A$ contains a zero column, since any linear combination of the rows will have a zero in that position. But more generally it cannot be the case if there is some non-zero (column) vector $v$ in the kernel of $A$, that is with $Av=0$, since any linear combination $C$ of the rows of $A$ will have $Cv=0$ as well, while there certainly exists a row $R$ of $I$ with $Rv\neq0$ (just take it at an index where $v$ has a non-zero coefficient). The case of a zero column in $A$ is just the one where $v$ can be taken to be a vector in the standard basis, but there really is not so much special about these vectors.

Now if $A$ has a zero row, then the remaining two rows only span a $2$-dimensional subspace$~W$ of the $3$-dimensional space of all possible rows, so there is bound to be some vector $v$ with $Rv=0$ for all rows $R\in W$; only, contrary to the case of a zero column, the given information is not sufficient to give an explicit such$~v$ (one would need to know the remaining two rows of$~A$ for this).

In general one can show, for square matrices$~A$, that the existence of a non-zero column$~v$ with $Av$is equivalent to the existence of a non-zero row $r$ with $rA=0$ (because row rank equals column rank), but although one or the other condition is obviously satisfied when there is a zero row or column, this is far from being the only case where the equivalent conditions are satisfied (and where therefore $A$ is not invertible).

Marc van Leeuwen
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