Answer to Question #156298 in Linear Algebra for Ange

Question

Let A = {(x, y, z) E R^3ㅣx - y + z = 0} where E represents "an element of" and R the set of real numbers
A mapping f is defined in R^(2*2) , the set of all square matrices of order 2 with entries in R by f: R^(2*2) → R^2
A → Av where v = (-1 2).
(a) Determine ker F. Give a basis of it and precise its dimension.
(b)Verify that the matrix B = (2 1
4 2)
belongs to ker F, and determine its coordinate vector with respect to the basis you gave in (a) above

Answer 1

(a) Let us find determine $\ker F$ and give a basis of it and precise its dimension.

$\ker F=\big\{\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)\in\mathbb R^{2*2}\ |\ \left(\begin{array}{cc} a & b \\ c & d \end{array}\right)\left(\begin{array}{cc} -1 \\ 2 \end{array}\right)= \left(\begin{array}{cc} 0 \\ 0 \end{array}\right)\big\} =\big\{\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)\in\mathbb R^{2*2}\ |\ \left(\begin{array}{cc} -a+2b \\ -c+2d \end{array}\right)= \left(\begin{array}{cc} 0 \\ 0 \end{array}\right)\big\}= \big\{\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)\in\mathbb R^{2*2}\ |\ a=2b, c=2d\big\} =\big\{\left(\begin{array}{cc} 2b & b \\ 2d & d \end{array}\right)\ |\ b,d\in\mathbb R\big\}$

It follows that $\ker F=\big\{\left(\begin{array}{cc} 2b & b \\ 2d & d \end{array}\right)\ |\ b,d\in\mathbb R\big\}=\big\{b\left(\begin{array}{cc} 2 & 1 \\ 0 & 0 \end{array}\right)+d\left(\begin{array}{cc} 0 & 0 \\ 2 & 1 \end{array}\right)\ |\ b,d\in\mathbb R\big\}$, and therefore the linear independent matrices $\left(\begin{array}{cc} 2 & 1 \\ 0 & 0 \end{array}\right),\left(\begin{array}{cc} 0 & 0 \\ 2 & 1 \end{array}\right)$ form a basis of $\ker F.$ Thus $\dim(\ker F)=2.$

(b) Since for the matrix $B =\left(\begin{array}{cc} 2 & 1 \\ 4 & 2 \end{array}\right)$ we have that $\left(\begin{array}{cc} 2 & 1 \\ 4 & 2 \end{array}\right)\left(\begin{array}{cc} -1 \\ 2 \end{array}\right)=\left(\begin{array}{cc} 0 \\ 0 \end{array}\right)$, $B$ belongs to $\ker F$. Taking into account that $B=\left(\begin{array}{cc} 2 & 1 \\ 4 & 2 \end{array}\right)=1\cdot\left(\begin{array}{cc} 2 & 1 \\ 0 & 0 \end{array}\right)+2\cdot \left(\begin{array}{cc} 0 & 0 \\ 2 & 1 \end{array}\right)$, we conclude that its coordinate vector with respect to the above basis is $(1,2).$