Question

Consider the linear transformation$\mathit{T}\mathbf{\left(}\overrightarrow{\mathbf{x}}\mathbf{\right)}\mathbf{=}\mathit{d}\mathit{e}\mathit{t}\left[\begin{array}{ccccc}1& \mid & & \mid & \mid \\ {\overrightarrow{v}}_1& {\overrightarrow{v}}_2& ...& {\overrightarrow{v}}_{n-1}& \overrightarrow{x}\\ \mid & \mid & & \mid & \mid \end{array}\right]$

From ${\mathit{R}}^{\mathbf{n}}$to$ℝ$ , whererole="math" localid="1668172293295" ${\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{..}\mathbf{,}{\overrightarrow{\mathbf{v}}}_{\mathbf{n}\mathbf{-}\mathbf{1}}$ , are linearly independent vectors in${\mathit{R}}^{\mathbf{n}}$ . Describe image and kernel of this transformation, and determine their dimensions.

Short answer

Therefore, the dimensions of kernel and image is given by,

$\ker T=\mathrm{span}\left({\overrightarrow{v}}_1,\overrightarrow{v_2},\dots ,v_{n-1}\right)$, $\mathrm{dimker}T=n-1$

$\mathrm{im}T=ℝ$,$\mathrm{dimim}T=1$

Step-by-step solution

Definition 

A determinant is a unique number associated with asquare matrix.

A determinant is a scalar value that is a function of the entries of a square matrix.

It is the signed factor by which areas are scaled by this matrix. If the sign is negative the matrix reverses orientation.

Given 

Given the linear transformation,

$T\left(\overrightarrow{x}\right)=\det \left[\begin{array}{ccccc}1& \mid & & \mid & \mid \\ {\overrightarrow{v}}_1& {\overrightarrow{v}}_2& \cdots & {\overrightarrow{v}}_{n-1}& \overrightarrow{x}\\ \mid & \mid & & \mid & \mid \end{array}\right]$

To determine the dimensions of kernel and image

For$\overrightarrow{x}$to be in the kernel of$T$, the given matrix must be non-invertible, and since${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,\dots ,\overrightarrow{v_{n-1}}$linearly independent, then$\overrightarrow{x}$must be a linear combination of${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,\dots ,v_{n-1}\to$.

Therefore,$\ker T=\mathrm{span}\left({\overrightarrow{v}}_1,\overrightarrow{v_2},\dots ,v_{n-1}\right)$, and $\mathrm{dimker}T=n-1$.

Since the given matrix can be invertible or non-invertible, its determinant can be different than zero or zero itself. Given the fact that the image of $T$ must be a subspace of $ℝ$, we conclude $\mathrm{im}T=ℝ$, and $\mathrm{dimim}T=1$.