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$A = \left(\begin{array}{crc} 1 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 3 \end{array}\right)$

What is the dimension of im(A)?

What is the dimension of ker(A)?

I know that the im(A) is the span of the column vectors of A. However, I don't know how to find the span of a vector column. I'm also confused about how to find the kernel since I think that involves a transformation matrix T but I only have the normal matrix A...

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To find the kernel, how about you try finding the kernel of the transformation associated to $A$? That is, let $T : \mathbb R^3 \to \mathbb R^3$ given by $T(x) = Ax$, and then find its kernel. – Patrick Da Silva Mar 7 '12 at 7:19
This question is directly related to your previous one. $\mathrm{im}(A)$ is precisely the set of vectors $v$ for which the equation $Ax=v$ has a solution! – Arturo Magidin Mar 7 '12 at 17:46
up vote 1 down vote accepted

The last column depends on the two others : $$ \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} = (1)\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + (-2) \begin{bmatrix} 0 \\ -1 \\ -1 \end{bmatrix}. $$ Therefore the span of the three columns is actually the span of the two first ones. But the two first ones are linearly independent, for if $$ \alpha \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + \beta \begin{bmatrix} 0 \\ -1 \\ -1 \end{bmatrix} = \begin{bmatrix} \alpha \\ -\beta \\ \alpha-\beta \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}, $$ then $\alpha = -\beta = 0$, which means $\alpha = \beta = 0$. Since they are linearly independent, we know that $$ \mathrm{Im}(A) = \left\{ \left. \begin{bmatrix} \alpha \\ -\beta \\ \alpha-\beta \end{bmatrix} \right| \, \alpha, \beta \in \mathbb R \right\}. $$ To find the kernel, you need to solve the equation $Ax = 0$, which means solving $$ \left( \begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & -1 & 2 & 0 \\ 1 & -1 & 3 & 0 \\ \end{array} \right) $$ as you have already probably learned. I leave the computations up to you.

Now the question asked for the dimensions. If you only want this information, then I could've stopped at the part where I noticed that the first two columns are linearly independent and that the third depended on the two first ; this means that $\dim(\mathrm{Im}(A)) = 2$, and since $3 = \dim(\mathrm{Im}(A)) + \dim(\mathrm{Ker} (A)) = 2 + \dim(\mathrm{Ker}(A))$, we find that $\dim(\mathrm{Ker}(A)) = 1$ by the nullity-rank theorem. If you don't want to use this theorem you could compute ; you're getting the same answer.

Hope that helps,

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So does that mean that ker(A) = $\begin{bmatrix} x\\ y\\ z \end{bmatrix} = c\begin{bmatrix} -1\\ 2\\ 1 \end{bmatrix}$? – Kyle V. Mar 7 '12 at 16:53
@StickFigs: that doesn't really make sense. The kernel of $A$ is a set of vectors; more precisely, a linear subspace of $\mathbb{R}^3$. But it is right that the vector $(-1,2,1)^t$ spans the kernel as a linear subspace, and that every vector in $\ker(A)$ is of the form $c\dot(-1,2,1)^t$, $c\in\mathbb{R}$. – InvisiblePanda Mar 7 '12 at 17:56

I think we are working over $\mathbb{R}$ here, or maybe $\mathbb{C}$? In any case, the kernel of $A$ is the solution set (it is a linear subspace of $\mathbb{R}^3$ / $\mathbb{C}^3$. I'll just assume you are working over the reals from now on) of the equation $Ax=0$, $x\in\mathbb{R}^3$. This requires no transformation matrix to compute.

But here you are only asked about the dimension of the image and kernel, respectively. Remember that the dimension formula (not sure if that's the right English term, correct me if it's wrong please) yields

$$ \dim_K (\mathrm{Im}(f)) + \dim_K(\mathrm{Ker}(f))=\dim_K(V), $$

where $f:V\to W$ is a $K$-linear mapping between $K$-vector spaces.

In your case, $A$ represents an $\mathbb{R}$-linear mapping $\mathbb{R}^3\to\mathbb{R}^3$, so $\dim_K(V)=\dim_{\mathbb{R}}(\mathbb{R}^3)=3$. So if you can compute the dimension of the image or kernel, you automatically get the other one by that formula above.

As you mentioned, the image is the span of the columns of $A$. You can easily see that the third column is a linear combination of the first two, and that the first two columns are linearly independent. This yields $\dim(\mathrm{Im}(A))=2$, hence $\dim (\mathrm{Ker}(A))=1$. (The image itself is then the span of the two vectors $x=(1,0,1)^{\top}, y=(0,-1,-1)^{\top}$ in $\mathbb{R}^3$, i.e. consists of all linear combinations of the form $\alpha x+\beta y$, where $\alpha,\beta\in\mathbb{R}$ are scalars.)

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"The dimension formula" you cited is called the "rank-nullity theorem" in English. – Patrick Da Silva Mar 8 '12 at 22:16
Thank you @Patrick! – InvisiblePanda Mar 9 '12 at 10:48

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