Question

Let T: R^5 tends to R^3 defined by T(a, b, c, d, e)=(a+2b+2c+d+e, a+2b+3c+2d-e, 3a+6b+8c+5d-e). Find the basis and the dimension of the kernel and the image of T

Answer 1

Solution: Given that , T(a, b, c, d, e)=(a+2b+2c+d+e, a+2b+3c+2d-e, 3a+6b+8c+5d-e).

This can be written as $T\left[\begin{array}{c}a\\b\\c\\d\\e\\\end{array}\right]=\left[\begin{array}{c}1&2&2&1&1\\1&2&3&2&-1\\3&6&8&5&-1\\\end{array}\right]\left[\begin{array}{c}a\\b\\c\\d\\e\\\end{array}\right]$

We write this as $T(v)=Av ~where~v=\left[\begin{array}{c}a\\b\\c\\d\\e\\\end{array}\right] and ~A=\left[\begin{array}{c}1&2&2&1&1\\1&2&3&2&-1\\3&6&8&5&-1\\\end{array}\right]$

To find ker(T):

The $ker(T)$ is the same as the null space of the matrix $A$ $(i.e. null A)$ . Therefore we find reduced row echelon form of the matrix $A$

$\therefore rref(A)=\left[\begin{array}{c}1&2&2&1&1\\1&2&3&2&-1\\3&6&8&5&-1\\\end{array}\right]=\left[\begin{array}{c}1&2&0&-1&5\\0&0&1&1&-2\\0&0&0&0&0\\\end{array}\right]$

Therefore rewriting the system in terms of free variables$~b,d,e$, we have

$\left[\begin{array}{c}a\\b\\c\\d\\e\\\end{array}\right]=\left[\begin{array}{c}-2b+d-5e\\b\\-d+2e\\d\\e\\\end{array}\right]=b\left[\begin{array}{c}-2\\1\\0\\0\\0\\\end{array}\right]+d\left[\begin{array}{c}1\\0\\-1\\1\\0\\\end{array}\right]+e\left[\begin{array}{c}-5\\0\\2\\0\\1\\\end{array}\right]$

The set$[(-2,1,0,0,0),(1,0,-1,1,0),(-5,0,2,0,1)]$ spans $Null A$ and it is linearly independent set, So the set {(-2,1,0,0,0),(1,0,-1,1,0),(-5,0,2,0,1)} forms a basis of $\mathbf{ker(T)}$

Dimension of ker(T) : We know that "The dimension of Null A is the number of free variables in the equation $Ax=0$ "

Here, there are two leading variables $a ~and ~c$ and three free variables $b,d,e$ .

Hence dim Null A =3

Hence dim [ ker (T)]=3

To find Image of T i.e. Im(T):

To identify vector that span Im(T), we identify the first and third columns as pivot columns(leading 1's) in rref(A). These columns are linearly independent and span col(A).

$\therefore Im(T)=Col(A)=\left[\begin{array}{c}1\\1\\3\\\end{array}\right],\left[\begin{array}{c}2\\3\\8\\\end{array}\right]$

$Let~B=Im(T)=\left[\begin{array}{c}1&2\\1&3\\3&8\\\end{array}\right]$

$rref(B)=\left[\begin{array}{c}1&0\\0&1\\0&0\\\end{array}\right]$

rewriting the rref(B) in vector matrix form

$\Rightarrow \left[\begin{array}{c}a\\b\\\end{array}\right]=a\left[\begin{array}{c}1\\0\\\end{array}\right]+b\left[\begin{array}{c}0\\1\\\end{array}\right]$

Therefore, the Im(T) is the xy plane.The set{(1,0),(0,1)}Span Null B and it is linearly independent set, So the set {(1,0),(0,1)} forms a basis of Im(T).

Dimension of Im(T) : We know that rank(A)=2 (non-zero rows of rref(A))

and we know that dim(Im(A)=rank(A)

$\mathbf{\therefore dim(Im)=2}$