I'm learning linear algebra and trying to find the image, the kernel, their dimensions and their bases of a linear map $ \varphi (a,b,c,d) = (3a+2b, b-c,2d-a) $. Also I'm supposed to find the matrix of the linear map in bases $B=$ {$(1,0,0,0),(1,1,0,0), (1,1,1,0), (1,1,1,1)$} and $C=$ {$(1,1,1), (0,1,1), (0,0,-1)$}.
My solution is this:
Created a matrix of the linear map and solved all linear equations for $0$.
$\begin{pmatrix}
3 & 2 & 0 &0\\
0&1 &-1 &0 \\
-1&0 &0 &2
\end{pmatrix}$
$\sim $
$\begin{pmatrix}
1 & 0 & 0 &2\\
0&1 &0 &3 \\
0&0 &1 &3
\end{pmatrix}$ $x1=2s$ , $x2 = -3s$ , $x3 = -3s$ , $x4=s$
Therefore the kernel of $ \varphi $ = $<s*(2,-3,-3,1)>$ and $B_{ker}=(2,-3,-3,1)$ and the dimension of the kernel $=1$. Since it's $\mathbb{R}^4\rightarrow \mathbb{R}^3$, the dimension of image has to be $3$. We have pivots in the first 3 colums, so we can say that $<(3,0,-1),(2,1,0), (0,-1,0)>$ is the image of $ \varphi $. And for bases of $\varphi$, we can take $(3,0,−1),(2,1,0),(0,−1,0)$, as they are linearly independent.
Is my solution correct?
However, I don't know how to find the matrix in the bases B and C. Any help with that? Thanks!
Best Answer
There are a couple of mistakes. First, the Reduced Row Echelon Form (RREF) of the matrix should be \begin{equation} \begin{pmatrix} 1 & 0 & 0 & -2 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 3 \end{pmatrix} \end{equation} It should be $-2$ in the corner, not $2$. But surprisingly, the basis for $\ker (\varphi)$ you found is actually correct (you must have made a second sign error in the computation which cancelled out with the first error).
This is correct.
This is an incorrect statement; this I think is more of a terminology mistake rather than a deep conceptual one. The proper statement is $\{(3,0,−1),(2,1,0),(0,−1,0)\}$ forms a basis for the image of $\varphi$. Recall that the image of a linear map is a subspace of $\mathbb{R^3}$, so it can't just consist of $3$ vectors, but a basis for a $3$-dimensional image consists of $3$ vectors.
By the way, for this particular example, there is a much easier way to determine a basis for the image of $\varphi$. You already mentioned that the image has dimension $3$. But notice that the target space $\mathbb{R^3}$ also has dimension $3$. Hence, $\text{image}(\varphi) = \mathbb{R^3}$. So there's a particularly obvious basis: $\{(1,0,0), (0,1,0), (0,0,1) \}$.
Lastly, to compute $[\varphi]_B^C$, the matrix of $\varphi$ with respect to the bases $B$ and $C$, what you have to do is for each vector $v \in B$, compute what $\varphi(v)$ is, and write it as a linear combination of vectors from $C$. The coefficients will then be the entries of the matrix
For example, the first vector in $B$ is $(1,0,0,0)$. So, now we have to evaluate $\varphi$ on this vector: \begin{align} \varphi(1,0,0,0) &= (1,0,-1) \\ &= (1,1,1) - (0,1,1) + (0,0,-1) \end{align} Notice that the coefficients are $1,-1,1$. So, the first column of $[\varphi]_B^C$ looks like \begin{pmatrix} 1 & \cdot & \cdot & \cdot\\ -1 & \cdot & \cdot & \cdot \\ 1 & \cdot & \cdot & \cdot \end{pmatrix}
The second vector of $B$ is $(1,1,0,0)$. Now, we compute again: \begin{align} \varphi(1,1,0,0) &= (5,1,-1) \\ &= 5 (1,1,1) -4 (0,1,1) + 2(0,0,-1) \end{align} So, the first two out of four columns of $[\varphi]_B^C$ look like: \begin{equation} \begin{pmatrix} 1 & 5 & \cdot & \cdot\\ -1 & -4 & \cdot & \cdot \\ 1 & 2 & \cdot & \cdot \end{pmatrix} \end{equation} I'll leave it to you to figure out what the last two columns are (follow the same process I did).