Dimensions of image and kernel of a $n \times n$ matrix

Question

Let $K$ be a field, $V$ a vector space over $K$ of a finite dimension $n=dim_K(V)$.

Let $f : V \rightarrow V$ be a $K$-linear map and $\mathfrak{B}$ an ordered basis of $V$ with

$$M_{f, \mathfrak{B}, \mathfrak{B}} = \begin{pmatrix} 0 & 1 & 1 & 1 & \cdots &1 \\ 0 & 0 & 1 & 1 &\cdots&1 \\ 0 & 0 & 0 & 1 & \cdots &1 \\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ &&&&&1\\ 0 & 0 & 0 & 0 & \cdots & 0 \end{pmatrix}$$

Side questions: Does this matrix have a name? Does the basis with the columns as vectors have a name?

I need to calculate the dimensions of the image and the kernel of $f$, I know how to do it with a completely given matrix and basis, but without it I have problems. Any hints welcome.

Answer 1

Hint:

The rank of a matrix is the number of non-zero rows after the matrix has been written as an upper triangular matrix by row reduction. This one is already in upper triangular form.

For the dimension of the kernel, use the rank-nullity theorem.

Answer 2

We can row-reduce the matrix to $$\begin{pmatrix}0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & & \vdots \\&&&\ddots&\\ \vdots & & & \ddots& 1 \\ 0 &\cdots & & & 0\end{pmatrix}$$ and in this form it is clear that the rank is $n-1$. Hence, $$\dim\operatorname{im} f=n-1,$$ and by the rank-nullity theorem, $$\dim\ker f=n-(n-1)=1.$$