MHB Solving Linear Transformations w/ Bases of Vector Spaces

[Chris L T521]

Here is the question:

Let $\{e_1, e_2, e_3\}$ be a basis for the vector space $V$ and $T: V \rightarrow V$ a linear transformation.

(a)
Show that if $f_1=e_1+e_2+e_3$, $f_2=e_1+e_2$, $f_3=e_1$ then $\{f_1,f_2,f_3\}$ is also a basis for $V$

(b)
Find the matrix $B$ of the $T$ with respect to the basis $\{e_1,e_2,e_3\}$ given that its matrix with respect to $\{f_1,f_2,f_3\}$ is
\[A = \begin{pmatrix}0 & 0 & 1\\ 0 & 1 & 1 \\ 1 & 1 & 1\end{pmatrix}\]

Here is a link to the question:

Let {e1, e2, e3} be a basis for the vector space V and T: V -> V a linear transformation.? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Chris L T521]

Hello george,

Since $\{e_1,e_2,e_3\}$ forms a basis for $V$, we see that they're linearly independent, i.e. for scalars $c_1,c_2,c_3\in\mathbb{R}$ (or any field of interest) $c_1e_1+c_2e_2+c_3e_3 = 0 \iff c_1=c_2=c_3=0$. Now, let $d_1,d_2,d_3\in\mathbb{R}$. We now consider the equation $d_1f_1+d_2f_2+d_3f_3 = 0$. Since $f_1=e_1+e_2+e_3$, $f_2=e_1+e_2$ and $f_3=e_1$, we see that
\[\begin{aligned} d_1f_1+d_2f_2+d_3f_3 = 0 &\implies d_1(e_1+e_2+e_3) + d_2(e_1+e_2) + d_3(e_3) = 0 \\ & \implies (d_1+d_2+d_3)e_1 + (d_1+d_2)e_2 + d_1e_3 = 0\end{aligned}\]
Since $\{e_1,e_2,e_3\}$ was a basis, we must now have that
\[\left\{\begin{aligned} d_1 + d_2 + d_3 &= 0\\ d_1 + d_2 &= 0 \\ d_1 &= 0\end{aligned}\right.\]
which clearly has the solution $d_1=d_2=d_3=0$. Therefore, $d_1f_1 + d_2f_2 + d_3f_3 = 0 \iff d_1=d_2=d_3=0$ and hence $\{f_1,f_2,f_3\}$ is a linearly independent set in $V$. Furthermore, $\mathrm{Span}\{f_1,f_2,f_3\} = \mathrm{Span}\{e_1,e_2,e_3\} = V$; hence, $\{f_1,f_2,f_3\}$ is also a basis for $V$.Next, we're told that in terms of the basis $\{f_1,f_2,f_3\}$, the linear transformation $T:V\rightarrow V$ is represented by the matrix
\[A = \begin{pmatrix}T(f_1) & T(f_2) & T(f_3)\end{pmatrix} = \begin{pmatrix}0 & 0 & 1\\ 0 & 1 & 1 \\ 1 & 1 & 1\end{pmatrix}\]
Since $f_1 = e_1 + e_2 + e_3$, $f_2 = e_1 + e_2$, and $f_3 = e_1$, we see that
\[T(f_1) = T(e_1) + T(e_2) + T(e_3),\quad T(f_2) = T(e_1) + T(e_2),\quad\text{and}\quad T(f_3) = T(e_1).\]
Comparing the columns of these two matrices, we have the system of equations
\[\left\{\begin{aligned}T(e_1) + T(e_2) + T(e_3) &= { \begin{pmatrix}0\\ 0 \\ 1\end{pmatrix}} \\ T(e_1) + T(e_2) &= {\begin{pmatrix} 0 \\ 1 \\ 1\end{pmatrix}} \\ T(e_1) &= {\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix} }\end{aligned}\right.\]
Solving this system yields that
\[T(e_1) = \begin{pmatrix}1\\ 1\\ 1\end{pmatrix},\quad T(e_2) = \begin{pmatrix}-1\\ 0 \\ 0\end{pmatrix},\quad\text{and}\quad T(e_3) = \begin{pmatrix}0 \\ -1 \\ 0\end{pmatrix}\]
Therefore, in terms of the basis $\{e_1,e_2,e_3\}$, the linear transformation $T:V\rightarrow V$ is represented by the matrix
\[B = \begin{pmatrix} T(e_1) & T(e_2) & T(e_3)\end{pmatrix} = \begin{pmatrix}1 & -1 & 0\\ 1 & 0 & -1\\ 1 & 0 & 0\end{pmatrix}\]I hope this makes sense!

 

[Petrus]

Here is the question:
Here is a link to the question:

Let {e1, e2, e3} be a basis for the vector space V and T: V -> V a linear transformation.? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

An alternativ way for a) is to check that that $$det \neq 0$$ but the way Chris L T521 is probably the best!

Basicly what I did Was put $$f_1,f_2,f_3$$ as a columne to a matrice and Then calculate the determinant! (You can se in picture which is $$f_1,f_2,f_3$$

Regards,
$$|\pi\rangle$$

 

[Deveno]

Alternative 3:

Because the set $\{f_1,f_2,f_3\}$ has 3 elements it suffices to show that this set spans $V$ (since $V$ has dimension 3 because of the size of the given basis).

We have:

$e_1 = f_3$
$e_2 = f_2 - f_3$
$e_3 = f_1 - f_2$

so $c_1e_1 + c_2e_2 + c_3e_3 = c_1(f_3) + c_2(f_2 - f_3) + c_3(f_1 - f_2)$

$= c_3f_1 + (c_2 - c_3)f_2 + (c_1 - c_2)f_3 \in \text{span}(\{f_1,f_2,f_3\})$.

***********
Before tackling the second question, it bears looking into what the change-of-basis matrix $P$ is:

$P = \begin{bmatrix}0&0&1\\0&1&-1\\1&-1&0 \end{bmatrix}$

If we call our bases B (for the $e$'s) and C (for the $f$'s), then:

$P([\mathbf{v}]_B) = [\mathbf{v}]_C$

The problem gives us $P^{-1}$ straight off the bat, it is:

$P^{-1} = \begin{bmatrix}1&1&1\\1&1&0\\1&0&0 \end{bmatrix}$

So the matrix we desire is:

$A' = P^{-1}AP$, which sends:

$[\mathbf{v}]_B \to [\mathbf{v}]_C \to [T(\mathbf{v})]_C \to [T(\mathbf{v})]_B$

Computing, we have:

$A' = P^{-1}AP = \begin{bmatrix}1&1&1\\1&1&0\\1&0&0 \end{bmatrix} \begin{bmatrix}0&0&1\\0&1&1\\1&1&1 \end{bmatrix} \begin{bmatrix}0&0&1\\0&1&-1\\1&-1&0 \end{bmatrix}$

$ = \begin{bmatrix}1&2&3\\0&1&2\\0&0&1 \end{bmatrix} \begin{bmatrix}0&0&1\\0&1&-1\\1&-1&0 \end{bmatrix}$

$ = \begin{bmatrix}3&-1&-1\\2&-1&-1\\1&-1&0 \end{bmatrix}$

Huh. I get a different matrix. One of us must be wrong, who is it?

Chris L T521 writes:

$T(f_3) = T(e_1) = \begin{pmatrix}1\\1\\1 \end{pmatrix}$

This is misleading...coordinates don't tell us which vector we have UNTIL we choose a basis. What we actually have is:

$T(e_1) = T(f_3) = f_1 + f_2 + f_3 = (e_1 + e_2 + e_3) + (e_1 + e_2) + e_3 = 3e_1 + 2e_2 + e_3$.

We should expect that in the B-basis, $A'$ should map (1,0,0) to (3,2,1). I think I am correct.

EDIT:

Chris L T521, the mistake you made is that your images for $T(e_j)$ are still in the C-basis, you need to convert these back to the B-basis.

 

[Chris L T521]

Alternative 3:

Because the set $\{f_1,f_2,f_3\}$ has 3 elements it suffices to show that this set spans $V$ (since $V$ has dimension 3 because of the size of the given basis).

We have:

$e_1 = f_3$
$e_2 = f_2 - f_3$
$e_3 = f_1 - f_2$

so $c_1e_1 + c_2e_2 + c_3e_3 = c_1(f_3) + c_2(f_2 - f_3) + c_3(f_1 - f_2)$

$= c_3f_1 + (c_2 - c_3)f_2 + (c_1 - c_2)f_3 \in \text{span}(\{f_1,f_2,f_3\})$.

***********
Before tackling the second question, it bears looking into what the change-of-basis matrix $P$ is:

$P = \begin{bmatrix}0&0&1\\0&1&-1\\1&-1&0 \end{bmatrix}$

If we call our bases B (for the $e$'s) and C (for the $f$'s), then:

$P([\mathbf{v}]_B) = [\mathbf{v}]_C$

The problem gives us $P^{-1}$ straight off the bat, it is:

$P^{-1} = \begin{bmatrix}1&1&1\\1&1&0\\1&0&0 \end{bmatrix}$

So the matrix we desire is:

$A' = P^{-1}AP$, which sends:

$[\mathbf{v}]_B \to [\mathbf{v}]_C \to [T(\mathbf{v})]_C \to [T(\mathbf{v})]_B$

Computing, we have:

$A' = P^{-1}AP = \begin{bmatrix}1&1&1\\1&1&0\\1&0&0 \end{bmatrix} \begin{bmatrix}0&0&1\\0&1&1\\1&1&1 \end{bmatrix} \begin{bmatrix}0&0&1\\0&1&-1\\1&-1&0 \end{bmatrix}$

$ = \begin{bmatrix}1&2&3\\0&1&2\\0&0&1 \end{bmatrix} \begin{bmatrix}0&0&1\\0&1&-1\\1&-1&0 \end{bmatrix}$

$ = \begin{bmatrix}3&-1&-1\\2&-1&-1\\1&-1&0 \end{bmatrix}$

Huh. I get a different matrix. One of us must be wrong, who is it?

Chris L T521 writes:

$T(f_3) = T(e_1) = \begin{pmatrix}1\\1\\1 \end{pmatrix}$

This is misleading...coordinates don't tell us which vector we have UNTIL we choose a basis. What we actually have is:

$T(e_1) = T(f_3) = f_1 + f_2 + f_3 = (e_1 + e_2 + e_3) + (e_1 + e_2) + e_3 = 3e_1 + 2e_2 + e_3$.

We should expect that in the B-basis, $A'$ should map (1,0,0) to (3,2,1). I think I am correct.

EDIT:

Chris L T521, the mistake you made is that your images for $T(e_j)$ are still in the C-basis, you need to convert these back to the B-basis.

Ah, after posting my answer, I had a slight feeling something was off with the matrix I came up with. Thanks a bunch for clarifying that!