# Linear transformation

I have $2$ questions here, please help me thanks!!

1. Find $2$ bases $a=(v_1, v_2, v_3)$ and $b=(w_1, w_2, w_3)$ of $\mathbb{R}^3$ such that both of them are non standard and yet the matrix of the identity map $I: \mathbb{R}^3 \to \mathbb{R}^3$ is the identity map.

2. Let $U$ be a $4$ dimensional vector space over $\mathbb{R}$ and suppose that $U=V+W$, where $\dim V=\dim W=2$. Further, assume that $\{v_1,v_2,w_1,w_2\}$ is a basis of $U$, where $\{v_1,v_2\}$ is a basis of $V$ and $\{w_1,w_2\}$ is a bsis of $W$. Let $T: U \to U$ be a linear operator such that

1. $T$ sends every vector in $V$ to a vector in $V$ and every vector in $W$ to a vector in $W$
2. $T$ sends every vector in $V$ to a vector in $W$ and every vector in $W$ to a vector in $V$

What can you say about the matrix of $T$ with respect to the basis $\{v_1,v_2,w_1,w_2\}$ in either of these situations.

I don't really understand the questions. Thanks for the help again.

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Ya,im aware that a=b, but i can just choose any bases? Soory, I;m quite weak in linear algebra. – Chloe Mar 7 '13 at 3:46
Take $a$ any non standard basis. Set $b:=a$. Then the matrix is necessarily $I_4$. – 1015 Mar 7 '13 at 3:47

Hints:

1) Prove that the matrix of $Id$ is $I_3$ the identity matrix if and only if $a=b$.

2) Think $2\times 2$ blocks. For instance, if $A$ is the matrix of $T$ with respect to the basis you took, then the upper-left $2\times 2$ block is the matrix of the restriction of $T$ to $V$, followed by the projection onto $V$. Very concretely, take your first vector $v_1$. In case i), you know that $T(v_1)$ is in $V$, ie $T(v_1)=\lambda_1v_1+\lambda_2v_2$. What does that tell you about the first column of $A$?

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Ref to 2) Elements in column A will be λ1 and λ2? – Chloe Mar 7 '13 at 3:57
@Chloe Yes, exactly. The first column is $(\lambda_1,\lambda_2,0,0)$. I'm sure you can take it from here. – 1015 Mar 7 '13 at 4:21
Then for the first Q can I prove in this way?I(v1)=1.w1+0.w2+0.w3,I(v2)=0.w1+1.w2+0.w3,I(v3)=0.w1+0.w2+1.w3, therefore v1=w1,v2=w2,v3=w3 and a=b. – Chloe Mar 7 '13 at 4:29
@Chloe Yes. 2.1) The first two columns are of the form $(.,.,0,0)$. Then with the same reasoning, you see that the third and fourth columns are of the form $(0,0,.,.)$. This gives to zero blocks upper-right and lower-left. – 1015 Mar 7 '13 at 4:32
@Chloe Sorry, I did not see your edit. Yes, for the first Q, you've just shown that if the matrix is $I_4$, then the two bases are equal. But note the question actually leads you to do the converse. Take $a=b$ and show the matrix is $I_4$. I just gave the equivalence in my answer because I thought it would lead you to a better understanding. – 1015 Mar 7 '13 at 4:36

Pick any basis $a$ and $b$ such that $a = b$. Its easy to see why: $I(v_1) = v_1$ so you want $v_1 = w_1$ to give $T(v_1) = w_1$. This means that the entries in the first column are $1, 0, 0, 0$ and if you do the same for $v_2$ and $v_3$, then the diagonal entries of the matrix of $I$ will be $1$ and all other entries will be $0$. For the first part to the second question, we must have $$T(v_1) = av_1 + bv_2\\ T(v_2) = cv_1 + dv_2\\ T(w_1) = ew_1 + fw_2\\ T(w_2) = gw_1 + hw_2$$ for some constants $a, b, \dotsc, h$. So the matrix is in the form $$\begin{pmatrix} A & O\\ O & B \end{pmatrix}$$ where $A, B, O$ are matrices with $A,B$ contain the constants and $O$ being the zero matrix. Similarly, for the second part the matrix must be in the form $$\begin{pmatrix} O & A\\ B & O \end{pmatrix}$$

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Why the second part to the second q is (OBAO)?? – Chloe Mar 7 '13 at 4:09
All vectors in $V$ go to $W$. So we have $T(v_1) = aw_1 + bw_2$ and $T(v_2) = cw_1 + dw_2$ for some constant $a, \dotsc, d$ using the basis for $W$. Since it is an ordered basis, the first two columns are $0, 0, a, b$ and $0, 0, c, d$. So you can clearly see that the top left $2 \times 2$ block is zero and the bottom left is some matrix $B$ which we don't know. Finding the right side of the matrix is similar. To answer these questions, all you need to know is how to find the matrix of a linear map (which is very easy if you know it). If you don't know how to do it, please review it. – Pratyush Sarkar Mar 7 '13 at 4:15
Okay thank you! – Chloe Mar 7 '13 at 4:32