1. ## Linear Matrix question

if A is any mxn matrix such that m > n, is it possible to find an nxm matrix B such that $BA=I_n$? why or why not?

the second question is identical just switch m<n.

I dont understand this question at all. any help?

2. Originally Posted by p00ndawg
if A is any mxn matrix such that m > n, is it possible to find an nxm matrix B such that $BA=I_n$? why or why not?
If $A$ is such a matrix so that $A^TA$ is invertible then it can be shown $B= (A^TA)^{-1}A^T$ satisfies $BA = I_n$.

For example, let $A=\begin{bmatrix}1&0\\0&1\\1&1 \end{bmatrix}$ and follow the computations above to convince yourself.

There is a similar formula for the second part as well.

3. Originally Posted by ThePerfectHacker
If $A$ is such a matrix so that $A^TA$ is invertible then it can be shown $B= (A^TA)^{-1}A^T$ satisfies $BA = I_n$.

For example, let $A=\begin{bmatrix}1&0\\0&1\\1&1 \end{bmatrix}$ and follow the computations above to convince yourself.

There is a similar formula for the second part as well.

can you explain the bolded part. What does the raised to the -1 do the transpose and A?

woops it didnt bold, look at the [b]'s placed. sorry.

4. Originally Posted by p00ndawg
can you explain the bolded part. What does the raised to the -1 do the transpose and A?

woops it didnt bold, look at the [b]'s placed. sorry.
oh nvm, I got it, now im just having trouble with the m<n 2nd one.

Should I just use the same assumption but pick a different matrix in which m<n? and solve it similarly, to see what I geT?

5. Originally Posted by p00ndawg
if A is any mxn matrix such that m > n, is it possible to find an nxm matrix B such that $BA=I_n$? why or why not?

the second question is identical just switch m<n.
The rank of B cannot be greater than m, nor can the rank of BA. So if m<n then BA cannot be equal to I_n.