Question

Consider the linear transformation\(T(\vec x) = det\left( {\begin{aligned}{*{20}{c}}1&\mid &{}&\mid &\mid \\{{{\vec v}_1}}&{{{\vec v}_2}}&{...}&{{{\vec v}_{n - 1}}}&{\vec x}\\\mid &\mid &{}&\mid &\mid \end{aligned}} \right)\)

From\({R^n}\)to\(\mathbb{R}\), where\({\vec v_1},...,{\vec v_{n - 1}}\), are linearly independent vectors in\({R^n}\). Describe image and kernel of this transformation, and determine their dimensions.

Short answer

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

\(\ker T = {\mathop{\rm span}\nolimits} \left( {{{\vec v}_1},\overrightarrow {{v_2}} , \ldots ,{v_{n - 1}}} \right)\), \(\dim \ker T = n - 1\).

\({\mathop{\rm im}\nolimits} T = \mathbb{R}\), \(\dim {\mathop{\rm im}\nolimits} T = 1\).

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,

\(T(\vec x) = \det \left( {\begin{aligned}{*{20}{c}}1&\mid &{}&\mid &\mid \\{{{\vec v}_1}}&{{{\vec v}_2}}& \cdots &{{{\vec v}_{n - 1}}}&{\vec x}\\\mid &\mid &{}&\mid &\mid \end{aligned}} \right)\)

To determine the dimensions of kernel and image

For \(\vec x\) to be in the kernel of \(T\), the given matrix must be non-invertible, and since \({\vec v_1},{\vec v_2}, \ldots ,\overrightarrow {{v_{n - 1}}} \)linearly independent, then \(\vec x\) must be a linear combination of \({\vec v_1},{\vec v_2}, \ldots ,{v_{n - 1}} \to \).

Therefore,\(\ker T = {\mathop{\rm span}\nolimits} \left( {{{\vec v}_1},\overrightarrow {{v_2}} , \ldots ,{v_{n - 1}}} \right)\), and \(\dim \ker T = n - 1\).

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 \(\mathbb{R}\), we conclude \({\mathop{\rm im}\nolimits} T = \mathbb{R}\), and \(\dim {\mathop{\rm im}\nolimits} T = 1\).