Question

Let A and B be two matrices of the same size, with $\mathit{A}\mathbf{\ne }\mathit{B}$, both in reduced row-echelon form. Show that$\mathbf{Ker}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{\ne }\mathbf{ker}\mathbf{\left(}\mathbf{B}\mathbf{\right)}$. Hint: Focus on the first column in which the two matrices differ, say, the kth columns${\overrightarrow{\mathbf{a}}}_{\mathbf{k}}$and${\overrightarrow{\mathbf{b}}}_{\mathbf{k}}$of A and B, respectively. Explain why at least one of the columns${\overrightarrow{\mathbf{a}}}_{\mathbf{k}}$and${\overrightarrow{\mathbf{b}}}_{\mathbf{k}}$fails to contain a leading 1. Thus, reversing the roles of matrices A and B if necessary, we can assume that${\overrightarrow{\mathbf{a}}}_{\mathbf{k}}$does not contain a leading 1. We can write${\overrightarrow{\mathbf{a}}}_{\mathbf{k}}$ as a linear combination of preceding columns and use this representation to construct a vector in the kernel of A. Show that this vector fails to be in the kernel of B. Use Exercises 86 and 87 as a guide.

Short answer

Let

$\mathrm{A}=\left[\begin{array}{cccccc}1& 0& 2& 0& 4& 0\\ 0& 1& 3& 0& 5& 0\\ 0& 0& 0& 1& 6& 0\\ 0& 0& 0& 0& 0& 1\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{cccccc}1& 0& 2& 0& 0& 4\\ 0& 1& 3& 0& 0& 5\\ 0& 0& 0& 1& 0& 6\\ 0& 0& 0& 0& 1& 7\end{array}\right]$

be two matrices of same size such that $\mathrm{A}\ne \mathrm{B}$.Then

$$\begin{gathered} \mathrm{Ker}\left(\mathrm{A}\right)=\mathrm{spam}{\left\{\left[\begin{array}{c}-2\\ -3\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ -5\\ 0\\ -6\\ 1\\ 0\end{array}\right]\right\}}_{\mathrm{and}}\mathrm{Ker}\left(\mathrm{B}\right)=\mathrm{spam}\left\{\left[\begin{array}{c}-2\\ -3\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ -5\\ 0\\ -6\\ -7\\ 1\end{array}\right]\right\} \\ \mathrm{Ker}\left(\mathrm{A}\right)\ne \mathrm{Ker}\left(\mathrm{B}\right) \end{gathered}$$

Step-by-step solution

 Mentioning concept

Consider the matrices

$\mathrm{A}=\left[\begin{array}{cccccc}1& 0& 2& 0& 4& 0\\ 0& 1& 3& 0& 5& 0\\ 0& 0& 0& 1& 6& 0\\ 0& 0& 0& 0& 0& 1\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{cccccc}1& 0& 2& 0& 0& 4\\ 0& 1& 3& 0& 0& 5\\ 0& 0& 0& 1& 0& 6\\ 0& 0& 0& 0& 1& 7\end{array}\right]$

Such that the matrices A and B are of same size and $\mathrm{A}\ne \mathrm{B}$.

Since the rank of the matrix A and matrix B is 4.

dim(ker(A))=2 and dim(ker(B))=2

 Finding the ker(A)

Consider$\mathrm{A}\overrightarrow{\mathrm{x}}=0$

$$\begin{gathered} \Rightarrow \left[\begin{array}{cccccc}1& 0& 2& 0& 4& 0\\ 0& 1& 3& 0& 5& 0\\ 0& 0& 0& 1& 6& 0\\ 0& 0& 0& 0& 0& 1\end{array}\right]\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ {\mathrm{x}}_3\\ {\mathrm{x}}_4\\ {\mathrm{x}}_5\\ {\mathrm{x}}_6\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right] \\ \Rightarrow {\mathrm{x}}_1+2{\mathrm{x}}_3+4{\mathrm{x}}_5=0 \\ {\mathrm{x}}_2+3{\mathrm{x}}_3+5{\mathrm{x}}_5== \\ {\mathrm{x}}_4+6{\mathrm{x}}_5=0 \\ {\mathrm{x}}_6=0 \end{gathered}$$

Since the pivot elements are in first, second, fourth, and sixth columns, therefore${\mathrm{x}}_3\mathrm{and}{\mathrm{x}}_5$are free variables.

Let${\mathrm{x}}_3={\mathrm{k}}_1\mathrm{and}{\mathrm{x}}_5={\mathrm{k}}_2$

$$\begin{gathered} \Rightarrow x_1=-2k_1-4k_2 \\ {x}_2=-3k_1-5k_2 \\ {x}_4=-6k_2 \\ {x}_5=7k_2 \\ \Rightarrow \left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\\ x_5\\ x_6\end{array}\right]=k_1\left[\begin{array}{c}-2\\ -3\\ 1\\ 0\\ 0\\ 0\end{array}\right]+\left[\begin{array}{c}-4\\ -5\\ 0\\ -6\\ -7\\ 1\end{array}\right] \end{gathered}$$

role="math" localid="1659933386540" $\Rightarrow \mathrm{Ker}\left(A\right)=\mathrm{spam}\left\{\left[\begin{array}{c}-2\\ -3\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ -5\\ 0\\ -6\\ -7\\ 1\end{array}\right]\right\}$

 Finding the ker(B)

Let$\mathrm{B}\overrightarrow{\mathrm{x}}=0$

$$\begin{gathered} \Rightarrow \left[\begin{array}{cccccc}1& 0& 2& 0& 0& 4\\ 0& 1& 3& 0& 0& 5\\ 0& 0& 0& 1& 0& 6\\ 0& 0& 0& 0& 1& 7\end{array}\right]\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ {\mathrm{x}}_3\\ {\mathrm{x}}_4\\ {\mathrm{x}}_5\\ {\mathrm{x}}_6\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right] \\ \Rightarrow {\mathrm{x}}_1+2{\mathrm{x}}_3+4{\mathrm{x}}_5=0 \\ {\mathrm{x}}_2+3{\mathrm{x}}_3+5{\mathrm{x}}_5== \\ {\mathrm{x}}_4+6{\mathrm{x}}_5=0 \\ {x}_5+7{\mathrm{x}}_6=0 \end{gathered}$$

Since the pivot elements are in first, second, fourth, and fifth columns, therefore${\mathrm{x}}_3\mathrm{and}{\mathrm{x}}_6$are free variables.

Let${\mathrm{x}}_3={\mathrm{k}}_1\mathrm{and}{\mathrm{x}}_6={\mathrm{k}}_2$

$$\begin{gathered} \Rightarrow {\mathrm{x}}_1=-2{\mathrm{k}}_1-4{\mathrm{k}}_2 \\ {\mathrm{x}}_2=-3{\mathrm{k}}_1-5{\mathrm{k}}_2 \\ {\mathrm{x}}_4=-6{\mathrm{k}}_2 \\ {\mathrm{x}}_5=-7{\mathrm{k}}_2 \\ \Rightarrow \left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ {\mathrm{x}}_3\\ {\mathrm{x}}_4\\ {\mathrm{x}}_5\\ {\mathrm{x}}_6\end{array}\right]={\mathrm{k}}_1\left[\begin{array}{c}-2\\ -3\\ 1\\ 0\\ 0\\ 0\end{array}\right]+{\mathrm{k}}_2\left[\begin{array}{c}-4\\ -5\\ 0\\ -6\\ -7\\ 1\end{array}\right] \end{gathered}$$

$\Rightarrow \mathrm{Ker}\left(\mathrm{B}\right)=\mathrm{spam}\left\{\left[\begin{array}{c}-2\\ -3\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ -5\\ 0\\ -6\\ -7\\ 1\end{array}\right]\right\}$

 Final Answer

$$\begin{gathered} \mathrm{Since},\mathrm{for}\mathrm{the}\mathrm{matrix}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{such}\mathrm{that}A\ne B \\ \mathrm{Ker}\left(\mathrm{A}\right)=\mathrm{spam}{\left\{\left[\begin{array}{c}-2\\ -3\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ -5\\ 0\\ -6\\ 1\\ 0\end{array}\right]\right\}}_{\mathrm{and}}\mathrm{Ker}\left(\mathrm{B}\right)=\mathrm{spam}\left\{\left[\begin{array}{c}-2\\ -3\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ -5\\ 0\\ -6\\ -7\\ 1\end{array}\right]\right\} \\ \mathrm{Ker}\left(\mathrm{A}\right)\ne \mathrm{Ker}\left(\mathrm{B}\right) \end{gathered}$$