For each of the following m ×n matrices, decompose the first standard basis vector 𝐞1=𝐰+𝐳 ∈ℝ^n,

### Part (a) Matrix: u005C( A u003D u005Cbegin{pmatrix} 1 u0026 u002D2 u0026 1 u005Cu005C 2 u0026 u002D3 u0026 2 u005Cend{pmatrix} u005C)u000Au000A#### Step

For each of the following m ×n matrices, decompose the first...

For each of the following $m \times n$ matrices, decompose the first standard basis vector $\mathbf{e}_1=\mathbf{w}+\mathbf{z} \in \mathbb{R}^n$, where $\mathbf{w} \in \operatorname{coimg} A$ and $\mathbf{z} \in \operatorname{ker} A$. Verify your answer by expressing $\mathbf{w}$ as a linear combination of the rows of $A$. (a) $\left(\begin{array}{lll}1 & -2 & 1 \\ 2 & -3 & 2\end{array}\right)$, (b) $\left(\begin{array}{rrr}1 & 1 & 2 \\ -1 & 0 & -1 \\ -2 & -1 & -3\end{array}\right)$, (c) $\left(\begin{array}{rrrr}1 & -1 & 0 & 3 \\ 2 & 1 & 3 & 3 \\ 1 & 2 & 3 & 0\end{array}\right)$, (d) $\left(\begin{array}{rrrrr}-1 & 1 & 1 & -1 & 2 \\ -3 & 2 & -1 & -2 & 0\end{array}\right)$.

For each of the following $m \times n$ matrices, decompose the first standard basis vector $\mathbf{e}_1=\mathbf{w}+\mathbf{z} \in \mathbb{R}^n$, where $\mathbf{w} \in \operatorname{coimg} A$ and $\mathbf{z} \in \operatorname{ker} A$. Verify your answer by expressing $\mathbf{w}$ as a linear combination of the rows of $A$.
(a) $\left(\begin{array}{lll}1 & -2 & 1 \\ 2 & -3 & 2\end{array}\right)$,
(b) $\left(\begin{array}{rrr}1 & 1 & 2 \\ -1 & 0 & -1 \\ -2 & -1 & -3\end{array}\right)$,
(c) $\left(\begin{array}{rrrr}1 & -1 & 0 & 3 \\ 2 & 1 & 3 & 3 \\ 1 & 2 & 3 & 0\end{array}\right)$,
(d) $\left(\begin{array}{rrrrr}-1 & 1 & 1 & -1 & 2 \\ -3 & 2 & -1 & -2 & 0\end{array}\right)$.

Key Concepts

Linear Combination Representation

Expressing a vector as a linear combination of a set of basis vectors or spanning elements is fundamental to linear algebra. In this context, verifying that the component of the vector in the coimage lies in the row space is done by writing it as a linear combination of the rows of A. This technique confirms that the decomposition aligns with the structural subspaces associated with the matrix A.

Direct Sum Decomposition

Direct sum decomposition refers to the process of writing a vector space as the sum of two complementary subspaces, typically in a way that the intersection of these subspaces is only the zero vector. In the context of linear maps, the domain can be decomposed as the direct sum of the kernel of A and a complementary subspace, here selected as the coimage or row space of A. This decomposition allows for the unique separation of any vector into components that are annihilated by A and those that are not.

Kernel (Nullspace) of a Matrix

The kernel of a matrix A, often called the nullspace, is the set of all vectors v such that A*v = 0. This subspace of the domain contains all the solutions to the homogeneous equation and is a key component in understanding the structure of the linear transformation represented by A. It plays a central role in the decomposition of vectors in the domain, as any vector can be written uniquely as the sum of a vector in the kernel and a vector in a complementary subspace.

Row Space (Coimage) of a Matrix

The row space of a matrix, which can be identified with the coimage of the linear transformation A, is the span of the rows of A. It represents the portion of the domain that 'contributes' to the transformation into the codomain. In decompositions of vectors in the domain, the component lying in the row space is the one that is effectively seen by the matrix A, and it can be expressed as a linear combination of the rows of A.

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