Question

If the kernel of a 5 x 4 matrix A consists of the zero vector only and if$A\overrightarrow{u}=A\overrightarrow{w}$for two vectors $\overrightarrow{u}and\overrightarrow{v}$in${\mathrm{ℝ}}^4$, then the vectors must be equal.

Short answer

The above statement is true.

If the kernel of a 5 x 4 matrix A consists of the zero vector only and if $A\overrightarrow{u}=A\overrightarrow{w}$for two vectors $\overrightarrow{u}\text{and}\overrightarrow{v}$in${\mathrm{ℝ}}^4$ , then the vectors$\overrightarrow{u}\text{and}\overrightarrow{v}$ must be equal.

Step-by-step solution

Mentioning the concept

Let A be a matrix of order 5 x 4.

Since it is given that kernel of the matrix A consists of only zero vector. Then

$\ker \left(A\right)=\overrightarrow{0}$

$\Rightarrow \dim \left(\ker \right(A\left)\right)=0$

$\Rightarrow \text{nullity of}A=0$

Then by Rank – Nullity theorem, we have

$Rank\left(A\right)+Nullity\left(A\right)=Dim\left(R^4\right)=4$

$\Rightarrow Rank\left(A\right)=4$

Therefore matrix A is invertible.

Thus, $A^{-1}$exists.

To prove Au→=Aw→ for two vectors u→andv→ in ℝ4

Since, for the matrix A of order 5 x 4, $A^{-1}$exists and it is given that $A\overrightarrow{u}=A\overrightarrow{w}$for two vectors$\overrightarrow{u}\text{and}\overrightarrow{v}$ in ${\mathrm{ℝ}}^4$, then we have

$A\overrightarrow{u}=A\overrightarrow{w}$

$\Rightarrow A^{-1}\left(A\overrightarrow{u}\right)=A^{-1}\left(A\overrightarrow{v}\right)$

$\Rightarrow \left(A^{-1}A\right)\overrightarrow{u}=\text{(}{A}^{-1}A)\overrightarrow{v}$

$\Rightarrow \overrightarrow{u}=\overrightarrow{v}\text{(}\because A^{-1}\text{A = I)}$

Final Answer

Since the kernel of a 5 x 4 matrix A consists of the zero vector only then matrix A is invertible and if$A\overrightarrow{u}=A\overrightarrow{w}$ for two vectors$\overrightarrow{u}\text{and}\overrightarrow{v}$ in${\mathrm{ℝ}}^4$ , then the vectors$\overrightarrow{u}\text{and}\overrightarrow{v}$ must be equal.