Answer to Question #232549 in Linear Algebra for Amuj

Question

Let  be a linear transformation. Let  and  be zero vectors of  and. Show that

Answer 1

$Let \ T: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m} \\$  be a linear transformation.

$Let \ \mathbf{0}_{n} \ and \ \mathbf{0}_{m} \ be \ zero \ vectors \ of \ \mathbb{R}^{n} \ and \ \mathbb{R}^{m} , respectively. \\Show \ that \ T\left(\mathbf{0}_{n}\right)=\mathbf{0}_{m}.$

$Observe \ that \ we \ have\ 0 \cdot \mathbf{0}_{n}=\mathbf{0}_{n} .\\ (This \ is \ a \ scalar \ multiplication \ of \ the \ scalar \ 0 and \ the \ vector \ \mathbf{0}_{n} \\ Now \ we \ have\ T\left(\mathbf{0}_{n}\right)=T\left(0 \cdot \mathbf{0}_{n}\right)=0 \cdot T\left(\mathbf{0}_{n}\right)=\mathbf{0}_{m}\\ Here \ we \ used \ the \ one\ of\ the \ properties \ of \ the\ linear\ transformation \ T \ \\ in \ the\ second \ equality. \\ Note \ that \ \mathbf{0}_{n}=\mathbf{0}_{n}-\mathbf{0}_{n} \\ Thus \ we \ have \\ T\left(\mathbf{0}_{n}\right)=T\left(\mathbf{0}_{n}-\mathbf{0}_{n}\right)=T\left(\mathbf{0}_{n}\right)-T\left(\mathbf{0}_{n}\right)=\mathbf{0}_{m}\\ where \ we \ used \ the \ linearity \ of \ T \ in \ the \ second \ equality.\\ In \ the \ last \ equality, \ note \ that \ the \ vector \ T\left(\mathbf{0}_{n}\right) \ is \ m -dimensional \ vector.$