# Finding the image, the kernel, their dimensions and their bases of a linear map - verify my solution

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 $$\varphi$$ = $$$$ and $$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!

There are a couple of mistakes. First, the Reduced Row Echelon Form (RREF) of the matrix should be $$$$\begin{pmatrix} 1 & 0 & 0 & -2 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 3 \end{pmatrix}$$$$ It should be $$-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).

and the dimension of the kernel $$=1$$. Since it's $$\mathbb{R^4} \to \mathbb{R^3}$$, the dimension of image has to be $$3$$.

This is correct.

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.

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: $$$$\begin{pmatrix} 1 & 5 & \cdot & \cdot\\ -1 & -4 & \cdot & \cdot \\ 1 & 2 & \cdot & \cdot \end{pmatrix}$$$$ I'll leave it to you to figure out what the last two columns are (follow the same process I did).

• Thanks for such a detailed answer. Your explanation was superb, now I fully understand how to compute the matrix with respect to the bases B and C. Also the mistake in reduced row of the matrix was just a typo when creating the question that somehow passed by. :) Jun 10 '19 at 17:12