# How do I find the matrix M that transforms a vector A into the vector B?

How can I find the matrix that transforms a vector A to a vector B ? By that I mean, if I have a vector A(x, y, z), what is the matrix M that transforms it to the vector B(a, b, c) ?

What I am trying to do is this; I have three vector A, B, C and I want to find the matrix M that transforms those vector to D, E, F (using only one matrix M).

-
What size are the vectors? – Arturo Magidin Nov 24 '11 at 22:17
I am using 3 components vectors (x, y, z) – dotminic Nov 24 '11 at 22:20
Can you find a matrix sending (1,0,0) to D; (0,1,0) to E; and (0,0,1) to F? Can you find a matrix sending A to (1,0,0); B to (0,1,0); and C to (0,0,1)? If so, you can compose these. – Jonas Kibelbek Nov 24 '11 at 22:27
Are your vectors orthogonal, means: are they a basis? – draks ... Nov 24 '11 at 22:33
@AndreasSpörl: pairwise orthogonal and nonzero implies they are a basis, but they can be a basis without being orthogonal. The right question is "Are they linearly independent?" – Arturo Magidin Nov 24 '11 at 22:52

with $X = \left[ A, B, C \right]$ and $Y = \left[ D, E, F \right]$ you are looking for $Y = M \cdot X$ ? Right? Then just do this $$M = Y \cdot X^{-1}$$

where $X$, $Y$ and $M$ are $3\times3$ matrices.

-

In general, the problem you describe is not well-defined.

For example, there are many matrices that will map $(1,1,1)$ to $(1,2,3)$: pick your favorite two vectors, $\mathbf{v}_1$ and $\mathbf{v}_2$, and there is a matrix that sends $(0,1,0)$ to $\mathbf{v}_1$, $(0,0,1)$ to $\mathbf{v}_2$, and sends $(1,1,1)$ to $(1,2,3)$.

For the second problem you describe, it depends on A, B, and C. If you are talking about 3-dimensional vectors, and A, B, and C are linearly independent, then you can always find one and only one matrix that will do it (I explain how below). But if they are linearly dependent, then it may be impossible, depending on D, E, and F. For example, if $A=(1,0,0)$, $B=(0,1,0)$, and $C=(1,1,0)$, then there is no matrix that sends $A$ to $(1,0,0)$, $B$ to $(0,1,0)$, and $C$ to $(0,0,1)$ (because $C=A+B$, so a matrix that sends $A$ to $D$ and $B$ to $E$ must send $C$ to $D+E$).

Assuming that $A$, $B$, and $C$ are 3-dimensional vectors, and that they are linearly independent (there is no way to pick scalars $a,b,c$, not all of them zero, such that $aA + bB + cC = (0,0,0)$, then what you want can always be done. Here is one way of doing it:

Construct the matrix that has $A$, $B$, and $C$ as its columns. Call it $T$.

Then find $T^{-1}$, the inverse of $T$ (if $A$, $B$, and $C$ are linearly independent, this can be done). Now let $R$ be the matrix that as $D$, $E$, and $F$ as its columns.

The matrix you want is $RT^{-1}$.

For example, say $A=(1,1,1)$, $B=(1,0,1)$, $C=(1,2,3)$. and $D=(3,1,-1)$, $E=(0,0,1)$, $F=(3,4,0)$. Then $T$ will be $$T=\left(\begin{array}{ccc} 1& 1 & 1\\ 1 & 0 & 2\\ 1 & 1 & 3 \end{array}\right).$$ We find its inverse: \begin{align*} \left(\begin{array}{ccc|ccc} 1& 1 & 1 & 1 & 0 & 0\\ 1 & 0 & 2 & 0 & 1 & 0\\ 1 & 1 & 3 & 0 & 0 & 1 \end{array}\right) &\to \left(\begin{array}{rrr|rrr} 1 & 1 & 1 & 1 & 0 & 0\\ 0 & -1 & 1 & -1 & 1 & 0\\ 0 & 0 & 2 & -1 & 0 & 1 \end{array}\right) \to \left(\begin{array}{rrr|rrr} 1 & 1 & 1 & 1 & 0 & 0\\ 0 & 1 & -1 & 1 & -1 & 0\\ 0 & 0 & 1 & -\frac{1}{2} & 0 & \frac{1}{2} \end{array}\right)\\ &\to\left(\begin{array}{rrr|rrr} 1 & 1 & 0 & \frac{3}{2} & 0 & -\frac{1}{2}\\ 0 & 1 & 0 & \frac{1}{2} & -1 & \frac{1}{2}\\ 0 & 0 & 1 & -\frac{1}{2} & 0 & \frac{1}{2} \end{array}\right)\\ & \to \left(\begin{array}{rrr|rrr} 1 & 0 & 0 & 1 & 1 & -1\\ 0 & 1 & 0 & \frac{1}{2} & -1 & \frac{1}{2}\\ 0 &0 & 1 & -\frac{1}{2} & 0 & \frac{1}{2} \end{array}\right),\end{align*} so $$T^{-1} = \left(\begin{array}{rrr} 1 & 1 & -1\\ \frac{1}{2} & -1 & \frac{1}{2}\\ -\frac{1}{2} & 0 & \frac{1}{2} \end{array}\right).$$

Then $$M= RT^{-1} = \left(\begin{array}{rrr} 3 & 0 & 3\\ 1 & 0 & 4\\ -1 & 1 & 0 \end{array}\right)\left(\begin{array}{rrr} 1 & 1 & -1\\ \frac{1}{2} & -1 & \frac{1}{2}\\ -\frac{1}{2} & 0 & \frac{1}{2} \end{array}\right).$$

Explanation. The matrix $T$ maps $(1,0,0)$ to $A$, $(0,1,0)$ to $B$, and $(0,0,1)$ to $C$. The matrix $R$ sends $(1,0,0)$ to $D$, $(0,1,0)$ to $E$, and $(0,0,1)$ to $C$.

That means that the matrix $T^{-1}$ maps $A$ to $(1,0,0)$, $B$ to $(0,1,0)$, and $C$ to $(0,0,1)$. When we multiply $RT^{-1}$, we are composing the functions. So $A$ is sent to $(1,0,0)$ by $T^{-1}$, and then to $D$ by $R$; in summary, $A$ is mapped to $D$. Similarly, $B$ is sent first to $(0,1,0)$ by $T^{-1}$ and then to $E$ by $R$. And finally, $C$ is sent to $(0,0,1)$ by $T^{-1}$, and to $F$ by $R$. That's why $M$ does what you want it to do.

Caveat. If $A$, $B$, and $C$ are not linearly independent, then you need $D$, $E$, and $F$ to satisfy the same linear relations as $A$, $B$, and $C$ do in order for you to be able to find some map that works.

So, what do you do if $A$, $B$, and $C$ are not linearly independent?

If $A$ is the zero vector, then you need $D$ to be the zero vector as well; if it is not, there is no matrix $M$. If $A=D=\mathbf{0}$, then you can ignore them (any matrix will send $\mathbf{0}$ to $\mathbf{0}$); discard them, and proceed as described below.

If $A$ is not zero, but $B$ is a scalar multiple of $A$, $B=\lambda A$, then you need $E=\lambda D$. If this is not the case, then there is no matrix $M$. If this is the case, then you can discard $B$ and $E$, because if you find a matrix that sends $A$ to $D$, it will automatically send $B=\lambda A$ to $E=\lambda D$.

If $A$ is not zero, $B$ is not a scalar multiple of $A$, then $C$ is a linear combination of $A$ and $B$, $C=\alpha A + \beta B$. Then you need $F=\alpha D+\beta B$. If this is not true, then there is no matrix $M$. If this is true, then you can discard $C$ and $F$, because any matrix that sends $A$ to $D$ and $B$ to $E$ will automatically send $C=\alpha A + \beta B$ to $\alpha D+\beta E=F$.

So, now we are in a situation where we have two vectors and two images. Call them $A$ and $B$, with desired images $D$ and $E$.

If $A$ and $B$ are still linearly dependent, then one of them will be a multiple of the other; check to see that $D$ and $E$ also satisfy that condition; if they do not, there is no matrix $M$. If they do, then discard and ignore the vector that is a multiple of the other.

So now you are down to either 1 or 2 vectors that are linearly independent. Add vectors from among $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ so that you get three vectors that are a basis. Then find a matrix $M$, as described above for the linearly independent case, that does exactly what you want with your original vectors, and does anything to the new vectors (for example, sends them to $(0,0,0)$).

-

When you say "transforms" I take it you are referring to multiplication: you want $MA=B$. Let $R$ be the matrix whose columns are $A,B,C$, and let $S$ be the matrix whose columns are $D,E,F$. Can you see that $MR=S$? And can you see how to solve that equation for $M$?

-
Yes, solving MR = S, I would do M = S/R, which involves finding the inverse of R. – dotminic Nov 24 '11 at 22:51
Please don't ever write $S/R$ when $R$ is a matrix, you'll give your teacher a heart attack. $SR^{-1}$ is the preferred expression. – Gerry Myerson Nov 25 '11 at 0:11
Yes sorry, I wrote it that way because I was comparing it to a simple equation of the form ab = c where I would solve for a like this: a = c / b. I should have used the correct form though. – dotminic Nov 25 '11 at 1:50