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Mathematics 6 Online
OpenStudy (anonymous):

help needed with linear algebra!

OpenStudy (anonymous):

OpenStudy (anonymous):

to show that it is a linear operator, we have to show that \[g(r\vec{y})=rg(\vec{y})\]

OpenStudy (anonymous):

or that \[g(\mathbf{x}+\mathbf{y})=g(\mathbf{x})+g(\mathbf{y}) \]

OpenStudy (anonymous):

we have, by definition,\[ g(\mathbf{y})=\mathbf{P}\cdot \mathbf{x}\\ g(\mathbf{y})=\left[\begin{matrix}a&0&0\\0&b&0\\0&0&c \end{matrix}\right]\cdot \mathbf{x}\\ \;\\ g(\mathbf{x}+\mathbf{y})=\mathbf{P}\cdot(\mathbf{x}+\mathbf{y})\\ \text{we know that matrix multiplication is distributive}\\ g(\mathbf{x}+\mathbf{y})=P\cdot \mathbf{x}+P\cdot \mathbf{y}\\ g(\mathbf{x}+\mathbf{y})=g(\mathbf{x})+g(\mathbf{y}) \] QED

OpenStudy (loser66):

@ lynn this time, no way to get just the answer, right ? It's a "show" problem. Good luck

OpenStudy (loser66):

\[\huge\color{red}{Amazing~Art~in~Avatar}\]\[\huge\color{purple}{Got~It}\]

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