Linear Algebra Question 4

I am having trouble with the following problem and would greatly appreciate any help. (I had trouble properly formatting my matrices; sorry about that.)

Let $T: \mathbb{R}^4 \rightarrow \mathbb{R}^4$ be the linear transformation with standard basis $A= \left[ \begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \\ \end{array} \right]$.

Let $B=\{e_1,e_2,e_3,v\}$ be a basis of $\mathbb{R}^4$, where the $e_i$ are the standard basis vectors.

a). Find a vector v such that $[T]_B = \left[ \begin{array}{cccc} 1 & 0 & 0 & 8\\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 5 \\ 0 & 0 & 0 & 2 \\ \end{array} \right]$. Prove your answer is correct.

b). Prove that the first three columns of $[T]_B$ do not depend on v.

c). Prove that there is no choice of v such that $[T]_B = \left[ \begin{array}{cccc} 1 & 0 & 0 & 8\\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 5 \\ 0 & 0 & 0 & 3 \\ \end{array} \right]$.

-
Where does $\,v\,$ in (a), (c) appears afterwards...?? Is it supossed the fouth column vector in those matrices or what? –  DonAntonio Mar 19 '13 at 17:28

a) This means $Tv=8e_1+3e_2+5e_3+2v$. Start out from the given matrix $A$ which means that $Te_4=2e_4$ and write $v=v_1e_1+v_2e_2+v_3e_3+v_4e_4$.

b) The $i$th column contains the coordinates of $Tb_i$ w.r.t. the given basis, where $b_i$ is the $i$th member of this basis. Since $Te_i=e_i$ for $i=1,2,3$ (knowing this from $[T]=A$), this is indeed independent from $v$.

c) The lower right member must be $2$, it will follow from the computation for a), but in an easier way, you can argue that $\det [T]_B$ is independent from the basis $B$ (also because $[T]_B=B^{-1}TB$ where the columns of matrix $B$ are the basis vectors in the original basis).

-

This is just a hint to get started:

For a and b: Note that you have $T(e_i) = e_i$ for $i=1,2,3$ and you have $Te_4 = 2e_4$. You want to find a vector $v = v_1e_1 + v_2e_2 + v_3e_3 + v_4e_4$ such that $$Tv = 8e_1 + 3e_2 + 5e_3 + 2v.$$ So you get $$v_1e_1 + v_2e_2 + v_3e_3 + 2v_4e_4 = 8e_1 + 3e_2 + 5e_3 + 2(v_1e_1 + v_2e_2 + v_3e_3 + v_4e_4).$$ Try to solve this for $v_1, v_2, v_3, v_4$ (you might have multiple possible values for $v_4$).

For $c$ try the same with: $$Tv = 8e_1 + 3e_2 + 5e_3 + 3v$$ Unless I am wrong this should give you that $3v_4 = 2v_4$. That would mean that $v_4 = 0$ and so $v$ would not be linearly independent from $\{e_1, e_2, e_3\}$.

-