# Linear transformation followed by change of basis

I have this problem: I've attempted calculating: $$[L(v_1)]_T$$ but cannot understand where I'm going wrong. Apparently the answer is (1, -1). Here is my attempt: Can anyone spot where I've gone wrong? I think I did something wrong in the transformation but I'm not sure why.

• You can typeset mathematics with MathJax. – Ennar Nov 6 '18 at 21:56
• $\;Lv_i\;$ is simply the $\;i\,-$ th column in the matrix representing $\;L\;$ wrt those basis: this is precisely the definition of matrix wrt basis! – DonAntonio Nov 6 '18 at 21:58
• @DonAntonio, true. But I'd say this is a clever exercise since knowing definition and understanding it are different things. Even more so if you don't really know definition precisely. – Ennar Nov 6 '18 at 22:12
• @Ennar And that is why I remarked that that is precisely how we define this thing. A careful student will go and read his notes. – DonAntonio Nov 6 '18 at 23:05

Hint: If $$A$$ is matrix of $$L$$ in pair of bases $$S$$ and $$T$$, to calculate $$Lv_1$$ using $$A$$ you first need to write $$v_1$$ in basis $$S$$.

• That's the thing that confuses me. Isn't $v_1$ a vector that is part of the basis S? Is it somehow saying to write the vector in terms of the basis it makes up? – dj1121 Nov 7 '18 at 0:39
• @dj1121, let me ask you, what do you think $[-1, 1, 0]$ stands for? – Ennar Nov 7 '18 at 0:47
• Ahh wait a minute. Ok so I know that the columns of the matrix representing the transformation are just the transformation applied to the column vectors in the basis. In this case they're coming from S but how should I know that they come from S? Why not from T? Is that because the basis S is a basis for R3 and the transformation goes from R3 -> R2? So that means [-1,1,0] is $v_1$ before you apply the transformation so it's $[v_1]_s$ right? – dj1121 Nov 7 '18 at 0:56
• @dj1121, let's step back a bit. $[-1,1,0]$ is just a convenient way to express a vector in a basis. Since it doesn't say otherwise, you should assume it's canonical basis $\{e_1,e_2,e_3\}$. Thus, $[-1,1,0]$ represents $-e_1 + e_2.$ But, your matrix accepts vector columns in basis $S$, not in canonical basis, so you multiplying $A\, [-1,1,0]$ calculates $L(-v_1+v_2)$, not $Lv_1$. What you need to do is find a column vector $[a,b,c]$ such that $av_1+bv_2+cv_3 = v_1$ and then calculate $A\, [a,b,c]$ to get $Lv_1$. – Ennar Nov 7 '18 at 1:01
• What I want to say is that you know that $v_1$ is represented by $[-1,1,0]$ in canonical basis. But when you give that column vector to matrix $A$, it doesn't know that you mean vector $v_1$, it thinks that $[-1,1,0]$ is in basis $S$, so for $A$ column $[-1,1,0]$ means $-v_1+v_2$. You have to write $v_1$ in a way that $A$ understands. – Ennar Nov 7 '18 at 1:10

HINT

In the given basis

$$L(v_1)=A\cdot v_1=[1,-1]=1\cdot w_1-1\cdot w_2$$

and so on.