Let $L_A$ :$R^{3}_{col}$ $\rightarrow$ $R^{3}_{col}$ , X $\rightarrow$ AX be operator of left multiplication by matrix $$A=\begin{bmatrix}1 & 2 & 1 \\ 0 & 1 & 1 \\ -1 & 3 &4\end{bmatrix}$$Find bases of :

a. kernel Ker $L_A$

b. image Im $L_A$

c. Ker $L_A$ + Im $L_A$ and Ker $L_A$ $\cap$ Im $L_A$

I found the null space of $L_A$ as [1,-1,1] and the image as [1,0,-1] and [2,1,3]. But I couldn't think of something for part c. Do we use the fact that dim Ker $L_A$ + dim Im $L_A$= dim $L_A$??

  • $\begingroup$ Assuming, your bases for Ker $L_A$ and Im $L_A$ are correct, then Ker $L_A$ + Im $L_A$ is spanned by those bases, i.e. by the three vectors [1,-1,1], [1,0,-1], [2,1,3]. Since we know that the second and third vector are linearly independent of each other, you only have to check if the first is a linear combination of those. $\endgroup$ – jflipp Nov 9 '14 at 17:31
  • $\begingroup$ Regarding a basis for Ker $L_A$ $\cap$ Im $L_A$, note that Ker $L_A$ $\cap$ Im\ $L_A$ = Ker $L_A|_{Im L_A}$. In order to find a basis for Ker $L_A|_{Im L_A}$, we put $$B := \begin{pmatrix} 1 & 2 \\ 0 & 1 \\ -1 & 3 \end{pmatrix},$$ i.e. the columns of $B$ are the basis vectors of Im $L_A$. Then $B \cdot \mathbb R^2 = Im L_A$. Now, find $U := Ker\ AB \subseteq \mathbb R^2$. Then $B\cdot U = Ker\ L_A \cap Im\ L_A$, and you just have to find a basis for that. $\endgroup$ – jflipp Nov 9 '14 at 17:38
  • $\begingroup$ What do you mean with the last part? How can i find the basis? @jflipp $\endgroup$ – abcdef Nov 12 '14 at 20:46
  • $\begingroup$ Could you please state the solutions clearer? I want to check it with my own answers. Thank you @jflipp $\endgroup$ – abcdef Nov 23 '14 at 22:16

Put $$ u = \begin{pmatrix}1 \\ 0 \\ -1 \end{pmatrix}, \quad v = \begin{pmatrix}2 \\ 1 \\ 3 \end{pmatrix}, \quad w = \begin{pmatrix}1 \\ -1 \\ 1 \end{pmatrix}. $$ We compute $Aw = 0$ and conclude that $w\in ker\ L_A.$ This implies $\dim\ ker\ L_A \geq 1$, since $w \neq 0,$ and further $\dim\ im\ L_A \leq 2$, since $3 = \dim\ im\ L_A + \dim\ ker\ L_A.$ Now, $u$ is the first column of $A$ and $v$ is the second column of $A$, so certainly $u,v \in im\ L_A.$ We also have that $u$ and $v$ are linearly independent. One way to see this is to note that the top left 2 by 2 determinant of $A$ is nonzero: $$ \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = 1. $$ This implies $\dim\ im\ L_A \geq 2,$ and so all in all we find $\dim\ im\ L_A = 2$ and $\dim\ ker\ L_A = 1.$ So indeed, $\{w\}$ is a basis of $ker\ L_A$ and $\{u,w\}$ is a basis of $im\ L_A.$

Next, we show that $u,v,w$ are all linearly independent. If this were not true, then $w$ would have to be a linear combination of $u$ and $v$, since $u$ and $v$ are linearly independent. So we write $w = \lambda u + \mu v$ and try to determine $\lambda$ and $\mu.$ By looking at the second coordinate we see that necessarily $\mu = -1,$ so we must have $w = \lambda u - v.$ This implies that $$ w + v = \begin{pmatrix}3 \\ 0 \\ 4 \end{pmatrix} $$ must be a multiple of $u$. But obviously, this is not true. So indeed $u,v,w$ are all linearly independent. This means that $ker\ L_A \cap im\ L_A = 0$ and $ker\ L_A + im\ L_A = \mathbb R^3_{col}.$ In particular, we can take $\{u,v,w\}$ as a basis for $ker\ L_A + im\ L_A.$


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