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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.

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I am not sure whether a linear transformation is equivalent to a matrix in your problem. If yes, Let the linear transformation be a matrix $T$. Then $TA=B, TC=D$ imply $T=BA^{-1}$ and $T=DC^{-1}$. Check whether $BA^{-1}=DC^{-1}$. If yes, then a unique linear transformation $T$ is determined. –  Shiyu Apr 22 '11 at 3:59
2  
Neither do I really understand the question nor do I understand @Shiyu's comment. I'd suggest thinking about the following question: What is the domain of $T$ and what is its range? What are their respective dimensions? Do $A$ and $C$ really form a basis of the domain? If yes, then the linear map $T$ is uniquely defined. If not, then it isn't. –  t.b. Apr 22 '11 at 4:52
    
I never claimed that the Transformation itself is a matrix, only that the input values are matrices. The other comment is most unhelpful. I understand all that, but am unable to answer the question. –  larry Apr 22 '11 at 4:55
2  
Ok, once again (despite your tone and I doubt you really understand as you claim): You don't specify the domain of $T$ at all, so what is it? Then you claim that $A$ and $C$ are a basis of the domain (so this would mean that the domain of $T$ is formed by the diagonal matrices) and in that case $T$ would be uniquely determined and of course linear. If the domain is formed by all $2 \times 2$-matrices, you'd have to extend $A$ and $C$ to a basis (by two other matrices $E$ and $F$, say) and you can choose the values of $T$ on $E$ and $F$ arbitrarily and extend linearly. –  t.b. Apr 22 '11 at 5:18
2  
@user9915: To the contrary, @Theo's comment was quite helpful. You posed an unclear question in which the domain wasn't specified; Theo suggested ways of clarifying it. I agree that you should answer precisely these two questions to clarify your question: What is the domain of $T$? And what do you mean by "two input matrices are bases"? That $A$ and $C$ together form a basis (not bases) of the domain? Or that $A$ and $C$ represent (not are) bases of the domain? –  joriki Apr 22 '11 at 5:35

2 Answers 2

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.

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You are half a minute ahead of me :) I guess basically we mean the same thing. –  GWu Apr 22 '11 at 5:22
    
@GWu: Almost -- my interpretations a) and b) fit into yours, but c) is different; it was an attempt to make sense of the statement "the two input matrices are bases" and to interpret the question such that the domain would be $2$-dimensional without explicitly restricting to the space spanned by $A$ and $B$. –  joriki Apr 22 '11 at 5:28
    
I guess "the two input matrices are bases" are his/her own words and don't think the original question(I'm assuming it's a from a book/homework) means this (c). –  GWu Apr 22 '11 at 5:32

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.

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