Find Basis of Image of a Matrix

Question

I'm very confused at the following question:

Find the basis for the image and a basis of the kernel for the following matrix: $\begin{bmatrix} 7 & 0 & 7 \\ 2 & 3 & 8 \\ 9 & 0 & 9 \\ 5 & 6 & 17 \end{bmatrix}$

I just don't know how to do any of this. We find the image by doing the following: $\begin{bmatrix} 7 & 2 & 9 & 5 \\\ 0 & 3 & 0 & 6\\ 7 & 8 & 9 &17 \end{bmatrix}$ Then, after doing RREF, we get: $\begin{bmatrix} 1 & 0 & \frac{9}{7} & \frac{1}{7} \\\ 0 & 1 & 0 & 2\\ 0&0&0&0 \end{bmatrix}$. This gives us an image of {$\begin{bmatrix} 1 \\ 0 \\ \frac{9}{7} \\ \frac{1}{7} \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \\ 2 \end{bmatrix}$}. However, I don't know how to proceed from here. Please help me?

Answer 1

So you have to find the basis for the image and kernel of the map $$T_A:\mathbb{R}^3\rightarrow \mathbb{R}^4:X\mapsto AX$$ where $$A=\begin{bmatrix} 7 & 0 & 7 \\ 2 & 3 & 8 \\ 9 & 0 & 9 \\ 5 & 6 & 17 \end{bmatrix}.$$

Im not a fan of blindly following an algorithm someone showed you. Instead, let's think about things nice and slowly.

So what belongs to $\text{im}(T_A)$? Suppose we take $X_1=\begin{bmatrix} 1\\0\\0\end{bmatrix}$. Then $T_A(X_1)=AX_1=\begin{bmatrix}7\\2\\9\\5 \end{bmatrix}$ belongs to the image by definition. Notice that this simply is the first column of $A$. Similarly, consider $X_2=\begin{bmatrix} 0\\1\\0\end{bmatrix},X_3=\begin{bmatrix} 0\\0\\1\end{bmatrix}$, then $T_A(X_i)$ yields the $i$-th column of $A$.

Clearly, $\text{im}(T_A)$ is generated by $T_A(X_1),T_A(X_2)$ and $T_A(X_3)$ (why?). So we have to check whether any of these vectors is redundant.

Just by eyeballing this, you can see that $T_A(X_3)=T_A(X_1)+2T_A(X_2)$. Thus $T_A(X_3)$ is redundant and clearly $T_A(X_1),T_A(X_2)$ are linearly independent. It follows that $\text{im}(T_A)$ has $\left\{T_A(X_1),T_A(X_2)\right\}$ as a basis.

So looking back at the above, we find the following general procedure: The image of $T_A$ is generated by the columns of $A$. If they are not linearly independent, we have to recursively throw away those columns that arise as linear combinations of the others. This exactly what you are doing when you look at the RREF of $A^T$.

Okay, so let's move to the kernel of $T_A$. First of all, general theory tells us that $\dim(\mathbb{R}^3)=\dim(\ker(T_A))+\dim(\text{im}(T_A))$. Hence $\dim(\ker(T_A))=1$. So we only have to find one non-zero vector $X$ such that $T_A(X)=0$. Now consider $X=\begin{bmatrix} a\\b\\c\end{bmatrix}=aX_1+bX_2+cX_3$. Then $T_A(X)=aT_A(X_1)+bT_A(X_2)+cT_A(X_3)$. We are looking for $X$ such that $T_A(X)=0$. We already knew that $T_A(X_3)=T_A(X_1)+2T_A(X_2)$ thus $X=\begin{bmatrix} 1\\2\\-1\end{bmatrix}\in \ker(T_A)$. Putting everything together, we find that $\begin{bmatrix} 1\\2\\-1\end{bmatrix}$ generates $\ker(T_A)$.

Answer 2

$Im(T)$ is the column span.

For vectors to be a basis they need to be:

1. linear independent
2. span the subspace

So we are left to check for linear dependence $$\begin{pmatrix} 7 & 0 & 7 \\ 2 & 3 & 8 \\ 9 & 0 & 9 \\ 5 & 6 & 17 \end{pmatrix}\approx \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

We have $2$ pivots in the in the first 2 columns so $$\{\begin{pmatrix} 7 \\ 2 \\ 9 \\ 5 \end{pmatrix},\begin{pmatrix} 0 \\ 3 \\ 0 \\ 6 \end{pmatrix}\}$$ are basis for the image

To find the kernel we need to solve $$\left(\begin{array}{ccc|c} 7 & 0 & 7 & 0\\ 2 & 3 & 8 & 0\\ 9 & 0 & 9 & 0\\ 5 & 6 & 17 & 0 \end{array}\right)$$

Answer 3

You're absolutely correct with your basis for the image of A, aka its row space. The basis for the kernel of A is found similarly: you must solve the homogeneous system $ A \mathbf x = \mathbf 0 $, where $\mathbf x$ in your case is the 3x1 column vector $\mathbf x = (x_1, x_2, x_3)^T $. The solution to this is found by row-reducing $A$ and solving: $$\begin{bmatrix} 7 & 0 & 7 \\ 2 & 3 & 8 \\ 9 & 0 & 9 \\ 5 & 6 & 17 \end{bmatrix} \to \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}.$$ So we have the system $$ x_1 + x_3 = 0 \\ x_2 + 2x_3 = 0.$$ Letting $x_3$ be a free variable, we obtain the vector $\mathbf x= (-x_3, -2x_3, x_3)^T = (-1, -2, 1)^T$ as the basis for the kernel of A. This answer makes sense, since the dimension of $\text{ker} A$ is equal to the number of free variables in the solution of the homogeneous system (i.e., $\text{dim ker} A = 1$).