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Consider $A=\pmatrix{1&2\\1&-2\\0&1}$.

I'm trying to see how I can find 2 different left inverses for this $3\times2$ matrix. If it has a left inverse, then it is injective.

Also, for the Right inverse, how come it doesn't have any?

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up vote 3 down vote accepted

You can find a left inverse of the form $\pmatrix{a&0&b\\c&0&d}$ simply by ignoring the middle row of $A$ and inverting the rest. Repeat that to find a left inverse of the form $\pmatrix{0&p&q\\0&r&s}$.

As for a right inverse, assume that $AB=I_{3\times 3}$. Then, in particular $AB\mathbf e_1=\mathbf e_1$, $AB\mathbf e_2=\mathbf e_2$, and $AB\mathbf e_3=\mathbf e_3$. But $\{B\mathbf e_1, B\mathbf e_2, B\mathbf e_3\}$ are three vectors in $\mathbb R^2$ and therefore they must be linearly dependent. Multiplying with $A$ cannot then stop them being linearly dependent. Therefore, by contradiction, $AB\ne I_{3\times 3}$.

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How did you come up with the conclusion of finding a left inverse by "ignoring the middle row of A". Can you demonstrate this further, or provide an example? Thanks – Buddy Holly Nov 5 '11 at 20:13
It comes naturally with a bit of experience with how matrix multiplication works. – Henning Makholm Nov 5 '11 at 20:26
@BuddyHolly A matrix is left invertible if and only if the number of columns= the number of linearly independent rows... You can try to prove this as a separate exercise :) Henning simply observed that by ignoring those rows, you get a square matrix with l.i. rows, thus invertible.... – N. S. Nov 6 '11 at 1:39

You need to solve:

$$\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 \\ 1 & -2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$

All matrices

$$ A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix} $$

satisfying the above equation will be left inverses. You can use the same argument to check whether any right inverses exist. However, this will not be the case since your matrix is not quadratic (if both exist they agree and the matrix must be quadratic).

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To solve this, we will just RREF it right? Then this will give us a system of euations to find a11, a12, a13, etc. to which I can fill in the 2 answers to which Henning said above? – Buddy Holly Nov 6 '11 at 4:42

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