Unique linear transformation

Question

Will values given below define a unique linear transformation? If so find the value of T for an arbitrary domain element. If not, give one of the formulas for T (if possible)?

A = matrix |2 0|
.................. |0 1|

B = matrix |4 3|

..................|0 2|

C = matrix |8 0|
..................|0 3|

D = matrix |1 5|

...................|0 7|

Q:

T[A] = [B] and T[C] = [D].

I'm not really sure how to set this up. I think it is because, if i'm not mistaken, the two input matrices are bases. But that's all I got -- and I'm not even sure about that. Also, please spend more time on the explanation of the steps on how to answer the question. I'm more concerned with that.

Answer 1

As stated in the comments, the question is somewhat unclear. I see three possible interpretations:

a) My first understanding of the question was that $T$ is an automorphism on the the vector space of $2\times2$ matrices, i.e. a linear transformation from that space to itself. In that case, the answer is "no", since that space is $4$-dimensional, so we'd need $4$ values and not $2$ to fix an automorphism.

b) If we stick with taking the matrices at face value as matrices, since both $A$ and $C$ are diagonal, another interpretation is that the domain is meant to be the space of all diagonal matrices. That space is $2$-dimensional, so the $2$ given values fix a linear transformation from it. Writing

$$E_1:=\left(\begin{array}{cc}1&0\\0&0\end{array}\right)$$

and

$$E_2:=\left(\begin{array}{cc}0&0\\0&1\end{array}\right)\;,$$

we can write the given data as

$$T(2E_1+E_2)=\left(\begin{array}{cc}4&3\\0&2\end{array}\right)$$

and

$$T(8E_1+3E_2)=\left(\begin{array}{cc}1&5\\0&7\end{array}\right)\;.$$

Then linearity and Gaussian elimination yields

$$T(c_1E_1+c_2E_2)=c_1\left(\begin{array}{cc}-5.5&-2\\0&0.5\end{array}\right)+c_2\left(\begin{array}{cc}15&7\\0&1\end{array}\right)\;.$$

c) Your somewhat unclear statement "the two input matrices are bases" suggests a third interpretation: We could interpret $A$ through $D$ as matrices representing bases of the intended domain $\mathbb R^2$ of $T$. (Note that "representing bases" and "being bases" are two quite different concepts.) The column vectors of the matrices could be intended to be taken as basis vectors of a basis, or, equivalently, the matrices could be intended to be taken as transforming from the basis to the canonical basis.

In this case, there is no such linear transformation, since the facts that the basis vector $\left(2\atop0\right)$ is transformed into $\left(4\atop0\right)$ and the basis vector $\left(8\atop0\right)$ is transformed into $\left(1\atop0\right)$ are incompatible. Indeed, under this interpretation, a single condition $T(A)=B$ (with non-singular $A$) would suffice to fix an automorphism on the domain.

Answer 2

To answer this question, we need to know which domain you are talking about.

If it's the whole space $\mathbb{M}^{2\times 2}$ of $2\times 2$ matrices, then the answer is no. Because the space $\mathbb{M}^{2\times 2}$ is of dimension $4$, and in order for a linear operator $T$ to be uniquely defined, we need to know how $T$ maps a basis of the space (i.e., $4$ linear independent vectors).

On the other hand, if the domain is the subspace of $\mathbb{M}^{2\times 2}$ spanned by the matrices $A$ and $C$, then the answer is yes. But I guess the question means the former.

The question is really misleading when expressed in terms of matrices. I think we should just think of it as vectors, that is, $$\mathbb{M}^{2\times 2}=\mathbb{R}^4, $$ and regard \begin{align} \begin{pmatrix} a&b\\ c&d \end{pmatrix} \text{ as } \begin{pmatrix} a\\ b\\ c\\ d \end{pmatrix}. \end{align} Then everything should be clear.

In particular, any linear transform is represented by a matrix, not $2\times 2$, but $4\times 4$. And to get a formula of $T$, you just solve the equations \begin{align} M_T\begin{pmatrix} 2\\ 0\\ 0\\ 1 \end{pmatrix}= \begin{pmatrix} 4\\ 3\\ 0\\ 2 \end{pmatrix}, M_T\begin{pmatrix} 8\\ 0\\ 0\\ 3 \end{pmatrix}= \begin{pmatrix} 1\\ 5\\ 0\\ 7 \end{pmatrix} \end{align} where $M_T$ is the matrix ($4\times 4$) for the liner transformation $T$. Note that there are $4\times 4=16$ unknowns and there are 8 equations. There should be infinitely many solutions and one should suffice.