The image is the set of all points in $\mathbb{R}^4$ that you get by multiplying this matrix to points in $\mathbb{R}^5$, you can find these by checking the matrix on the standard basis. The kernel is the set of all points in $\mathbb{R}^5$ such that, multiplying this matrix with them gives the zero vector. Again you can find this in a similar way.

Dec 5, 2019 · From what I've seen I should reduce the matrix in echelon form (which it already is), and pick the columns that are linearly independent. The zero vector is independent to itself so the basis I have for the image is the span of $$\begin{bmatrix}0\\0\\0\end{bmatrix}$$

$\begingroup$ @Nick: You have to be careful: yes, every matrix describes a linear transformation relative to the standard basis, but also every matrix describes a linear transformation relative to whatever bases you want to specify ahead of time. You need to know what "language" the matrix expects to hear, and what "language" the matrix is ...

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

Concerning your second question: Since the matrix has rank 3, you need three linearly independent column vectors of the matrix as a basis for the image. You can take the first three vectors, since they are linearly independent. Therefore, a basis of the image is $$(1, 0, -1, -1), \qquad (0, 2, -1, 0), \qquad (0, 1, 0, 1).$$

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Trying to understand matrix image. Ask Question Asked 11 years, 2 months ago. Modified 11 years, 2 months ago.

Apr 7, 2020 · 2.The column index of the first non zero value in each row equals to an image index in the original matrix A: img = {{1,2,3}, {3,4,5}} which would be our solution.

May 31, 2018 · The image of the matrix representation is $(\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix})$ My question is, how do I find the kernel and the image of the linear operator, when I know the Null space and the image of the corresponding matrix representation?

Nov 19, 2015 · Does computing the column space, range, and image of a matrix all produce the same answer? And are they all written the same way? for example. Basis of the column space = {v1,v2} Basis of the ima...