- #1

mathmari

Gold Member

MHB

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Let $\lambda\in \mathbb{R}$ and \begin{equation*}a=\begin{pmatrix}1 & 2 &-1& \lambda & -\lambda \\ 0 & 1 & -1& \lambda & 2\\ 2 & 2 & 1 & 1 & 3\lambda-1 \\ 1 & 1 & 1 & \lambda & 5\end{pmatrix}\in \mathbb{R}^{4\times 5}\end{equation*}

(a) Let $\lambda=1$. Determine a Basis $\mathcal{B}$ of $\mathbb{R}^5$ and a Basis $\mathcal{C}$ of $\mathbb{R}^4$, such that $\mathcal{M}_{\mathcal{C}}^{\mathcal{B}}(\phi_a)$ at the upper left side there is an unit matrix and elsewise zero.

(b) Determine the rank of $a$ and $a^T$.

For question (a) :

With $\lambda=1$ we get the matrix \begin{equation*}a=\begin{pmatrix}1 & 2 &-1& 1 & -1 \\ 0 & 1 & -1& 1 & 2\\ 2 & 2 & 1 & 1 & 2 \\ 1 & 1 & 1 & 1 & 5\end{pmatrix}\end{equation*}

Let $\mathcal{B}=\{b_1 , b_2, b_3, b_4, b_5\}$ and $\mathcal{C}=\{c_1 , c_2, c_3, c_4\}$.

It holds that \begin{equation*}\mathcal{M}_{\mathcal{C}}^{\mathcal{B}}(\phi_a)=\left (\gamma_{\mathcal{C}}(\phi_a(b_1))\mid \gamma_{\mathcal{C}}(\phi_a(b_2))\mid \gamma_{\mathcal{C}}(\phi_a(b_3)) \mid \gamma_{\mathcal{C}}(\phi_a(b_4)) \mid \gamma_{\mathcal{C}}(\phi_a(b_5))\right )\end{equation*}

It holds also that $\gamma_{\mathcal{C}}(\phi_a(b_i))=\begin{pmatrix}\alpha_1 \\ \alpha_2\\ \alpha_3\\ \alpha_4\end{pmatrix}$ with $\phi_a(b_i)=\alpha_1\cdot c_1+\alpha_2\cdot c_2+\alpha_3\cdot c_3+\alpha_4\cdot c_4$.

That $\mathcal{M}_{\mathcal{C}}^{\mathcal{B}}(\phi_a)$ at the upper left side there is an unit matrix and elsewise zero, does it mean it must be of the form \begin{equation*}a=\begin{pmatrix}1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0& 0 & 0\\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0\end{pmatrix}\end{equation*} ? Or what does this mean?:unsure: