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Trying to figure something general regarding row equivalent matrices.

Say I Have $A, B$ two matrices of some size $m \times n$. I know that $A, B$ are row equivalent and their rows are linearly independent. (e.g., $A, B$ rows are both a different span base for the same subspace).

So finally, there is a matrix $P$, invertible, such that $A = PB$.

Would like to determine a formula for $P$.

Seems that $P$ rows are: the coordinates of $B$ rows regarding $A$ base.

Is it correct?

How can this be proven?

Will this work if $A, B$ rows are not linearly independent?

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1 Answer 1

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Seems that $P$ rows are: the coordinates of $B$ rows regarding $A$ base.

Close: it's the other way around. That is, the rows of $P$ are the coordinates of $A$ with respect to the basis formed by the rows of $B$.

The first thing to note is that $$\begin{bmatrix} x_1 & x_2 & \ldots & x_n\end{bmatrix}\begin{bmatrix}v_1 \\ \hline v_2 \\ \hline \vdots \\ \hline v_n \end{bmatrix} = [x_1 v_1 + x_2 v_2 + \ldots + x_nv_n],$$ where $x_1, \ldots x_n$ are scalars and $v_1, \ldots, v_n$ are row vectors of the same dimension. That is, multiplying on the left by a row vector simply produces a linear combination of the rows, where the coefficients are just the entries in the row vector. Further, $$\begin{bmatrix}u_1 \\ \hline u_2 \\ \hline \vdots \\ \hline u_n \end{bmatrix}B = \begin{bmatrix}u_1B \\ \hline u_2B \\ \hline \vdots \\ \hline u_nB \end{bmatrix}.$$ That is, multiplying two matrices together is equivalent to multiplying each individual row of the first matrix to the second, and placing the resulting row vectors as rows in a new matrix. Both of these facts can be verified by the definition of matrix multiplication.

So, we can interpret the rows of $PB$ as linear combinations of the rows of $B$. That is, the rows of the resulting matrix $A$ are coordinate row vectors in terms of the rows of $B$. That is, the rows of $P$ are the coordinates of the rows of $A$ with respect to the rows of $B$.

Would like to determine a formula for $P$.

If you multiply both sides of $A = PB$ by $B^*$ ($B^\top$ is fine in the real case), then $B$ having linearly independent rows implies $BB^*$ is invertible. So, you can simply take $P = (B^*B)^{-1}A$.

Will this work if $A,B$ rows are not linearly independent?

You can get a formula for $P$ using the Moore-Penrose pseudoinverse. If $A = PB$, then using the pseudoinverse property $BB^+B = B$, we have $$AB^+B = PBB^+B = PB = A.$$ I claim that $$Q = AB^+ + W(I - BB^+)$$ is a viable solution to $P$, for any square matrix $W$ of the appropriate size (yes, there are indeed multiple solutions). We then have $$QB = (AB^+ + W(I - BB^+))B = AB^+B + W(B - BB^+B) = AB^+B + 0 = A.$$ So, every value of $Q$ is a suitable choice for $P$. Note that, when $B$ has linearly independent rows, then $BB^+ = I$, so there will be only one solution generated, which must be the unique solution.

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