Calculate the image and a basis of the image (matrix)
What's the image of the matrix? What's the basis of the image? $M=\begin{pmatrix} -1 & 1 & 1\\ -2 & -3 & 6\\ 0 & -1 & 1 \end{pmatrix}$
First transposed the matrix:
$M^{T}=\begin{pmatrix} -1 & -2 & 0\\ 1 & -3 & -1\\ 1 & 6 & 1 \end{pmatrix}$
Now we use Gauss and get zero lines. Take the first line and add it to the third:
$M^{T}=\begin{pmatrix} -1 & -2 & 0\\ 0 & -5 & -1\\ 1 & 6 & 1 \end{pmatrix}$
Take the first line and add it to the third:
$M^{T}=\begin{pmatrix} -1 & -2 & 0\\ 0 & -5 & -1\\ 0 & 4 & 1 \end{pmatrix}$
Multiply the second line with $4$, multiply the third line with $5$, then add second line to third:
$M^{T}=\begin{pmatrix} -1 & -2 & 0\\ 0 & -20 & -4\\ 0 & 0 & 1 \end{pmatrix}$
Transpose back:
$M=\begin{pmatrix} -1 & 0 & 0\\ -2 & -20 & 0\\ 0 & -4 & 1 \end{pmatrix}$
The image of the matrix is $\text{Im(M)}= \text{span} \left ( \left\{ \begin{pmatrix} -1\\ -2\\ 0 \end{pmatrix}, \begin{pmatrix} 0\\ -20\\ 4 \end{pmatrix},\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix} \right\} \right)$
The basis of the image is $\left\{ \begin{pmatrix} -1\\ -2\\ 0 \end{pmatrix}, \begin{pmatrix} 0\\ -20\\ 4 \end{pmatrix},\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix} \right\}$
Please tell me if I did everything correctly? It's very important for me to know as I would do it like that in the exam :)
I hope it's correct and please also tell me if the notation is.
$\endgroup$1 Answer
$\begingroup$The image of a matrix is the same as its column space. To find column space, you first find the row echelon form of the given matrix (do not transpose it). The definition of row-echelon form is:
- Rows with all zero's are below any nonzero rows
- The leading entry in each nonzero row is a one
- All entries below each leading "1" are zero
With the matrix in row-echelon form, the image (and column space) basis of the matrix comprises of the columns that contain a leading 1.
It is also useful to note that the dimensions (dim) of im(M) = dim(colM) = rank of M
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