Math 301 | Linear Algebra | Fall 2026

Let \(V\) be a vector space, \(\bu\) a vector in \(V\), and \(k\) a scalar. Then

1. \(0 \bu=\bz\)
2. \(k \bz = \bz\)
3. \((-1) \bu = -\bu\)

Let \(V\) be a vector space and let \(W\) be a subset of \(V\). Then \(W\) is a subspace of \(V\) if the following three conditions are satisfied:

1. for every \(\bu\) and \(\bv\) in \(W\), \(\bu+\bv\) is in \(W\) (i.e., \(W\) is closed under vector addition);
2. for every \(\bu\) in \(W\) and scalar \(k\), \(k\bu\) is in \(W\) (i.e., \(W\) is closed under scalar multiplication);
3. the zero vector of \(V\) lies in \(W\).

Let \(V\) be a vector space and let \(\bv_1,\ldots,\bv_r\) be vectors in \(V\). Then \(\sspan\{\bv_1,\ldots,\bv_r\}\) is a subspace of \(V\).

A set \(\{\bv_1,\ldots,\bv_r\}\) of two or more vectors is linearly dependent if and only if at least one of the vectors is a linear combination of the others.

Let \(\cB=\{\bv_1,\ldots,\bv_n\}\) be a basis for a vector space \(V\).

1. If any vector of \(V\) is added to \(\cB\), then \(\cB\) is no longer linearly independent.
2. If any vector is removed from \(\cB\), then \(\cB\) no longer spans \(V\).

Let \(\cB=\{\bv_1,\ldots\bv_n\}\) be a basis of a vector space \(V\). Then every \(\bu\) in \(V\) can be written in exactly one way as a linear combination of \(\bv_1,\ldots, \bv_n\), that is, can be expressed as \begin{equation*} \bu=c_1\bv_1+\cdots c_n\bv_n, \end{equation*} for unique scalars \(c_1,\ldots,c_n\).

All bases of a vector space \(V\) have the same number of elements.

In \(\bR^n\), the following have the same solutions:

1. The vector equation \(x_1\bv_1+\cdots +x_p\bv_p=\bu\).
2. The linear system of equations with augmented matrix \([\bv_1\cdots \bv_p\mid \bu]\).
3. The matrix equation \([\bv_1\, \cdots \bv_p]\,\bx=\bu\).

Let \(A\) be an \(m\times n\) matrix. Then

1. \(\nul A\) is a subspace of \(\bR^n\).
2. \(\col A\) is a subspace of \(\bR^m\).

Let \(A\) be a matrix with \(n\) columns. Then \({\dim (\nul A) + \dim (\col A) = n}\).

Let \(T:V\to W\) be linear. Then

1. \(T(\bz)=\bz\).
2. \(T(c_1\bv_1+\cdots +c_p\bv_p)=c_1T(\bv_1)+\cdots+c_pT(\bv_p)\), for any scalars \(c_1,\ldots,c_p\) and vectors \(\bv_1,\ldots,\bv_p\) in \(V\).

A matrix transformation is linear.

Let \(T:V\to W\) be linear. Then \(\kernel(T)\) is a subspace of \(V\) and \(R(T)\) is a subspace of \(W\).

Let \(A\) be an \(m\times n\) matrix, and let \(T:\bR^n\to \bR^m\) be the matrix transformation \(T(\bx)=A\bx\). Then \(\kernel(T)=\nul A\) and \(R(T)=\col A\).

Let \(T:V\to W\) be linear. Then \(\dim(\ker T) + \dim (R(T)) = \dim V\).

Let \(T:V\to W\) be linear.

1. \(T\) is one-to-one if and only if \(\ker T=\{\bz\}\).
2. \(T\) is onto if and only if \(R(T) = W\).

Let \(W\) be a subspace of \(V\). If \(\dim W=\dim V\), then \(W=V\).

Let \(T:V\to W\) be linear, and suppose that \(\dim V=\dim W\). Then \(T\) is one-to-one if and only if \(T\) is onto.

Let \(T:U\to V\) and \(S:V\to W\) be linear transformations. Then the composition \(S\circ T:U\to W\) is a linear transformation.

Let \(T:V\to W\) be a linear transformation. Then \(T\circ I=I\circ T=T\).

Let \(T:V\to W\) be one-to-one. Then \(T^{-1}\circ T=I\).

If \(T:V\to W\) is an isomorphism, then \(\dim V=\dim W\).

Suppose that \(V\) is a vector space and \(B=\{\bv_1,\ldots,\bv_n\}\) is a basis for \(V\). Then the mapping \(T:V\to \bR^n\) given by \(T(\bu)=[\bu]_B\) is an isomorphism.

Let \(T:\bR^n\to \bR^m\) be a mapping. Then \(T\) is a linear transformation if and only if \(T\) is a matrix transformation.

Let \(T:\bR^n\to \bR^m\) be a linear transformation. The standard matrix for \(T\) is \([T(\be_1)\cdots T(\be_n)]\).

Suppose that the standard matrix for \(S\) is \(A\) and the standard matrix for \(T\) is \(B\). Then the standard matrix for \(S\circ T\) is \(AB\).

Let \(T:\bR^n\to \bR^n\) be a linear transformation, and let \(A\) be the standard matrix for \(T\). Then \(T\) is an isomorphism if and only if \(A\) is invertible. In this case, the standard matrix for \(T^{-1}\) is \(A^{-1}\).

Let \(A\) be an \(n\times n\) matrix. Then \(A\) is invertible if and only if \(A\) can be row reduced to \(I_n\).

Let \(A\) be an \(n\times n\) matrix, and let \(\bb\) be a vector in \(\bR^n\). If \(A\) is invertible, then \(A\bx=\bb\) has a unique solution, namely, \(\bx=A^{-1}\bb\).

Let \(A\) be an \(n\times n\) matrix, and let \(T\colon \bR^n\to \bR^n\) be given by \(T(\bx)=A\bx\). The following are equivalent:

1. \(T\) is an isomorphism.
2. \(T\) is one-to-one.
3. \(T\) is onto.
4. \(\ker T = \{\bz\}\).
5. \(R(T) = \bR^n\).
6. \(A\) is invertible.
7. \(A\) row-reduces to \(I_n\).
8. \(\nul A = \{\bz\}\).
9. The columns of \(A\) are linearly independent.
10. \(\col A = \bR^n\).
11. The columns of \(A\) span \(\bR^n\).

Let \(T:V\to W\) be linear. Let \(\cB=\{\bu_1,\ldots,\bu_n\}\) be a basis for \(V\) and \(\cB'=\{\bw_1,\ldots,\bw_m\}\) a basis for \(W\). Then there exists a matrix \([T]_{\cB',\cB}\) such that for every \(\bv\) in \(V\), \([T(\bv)]_{\cB'}=[T]_{\cB',\cB}\cdot [\bv]_{\cB}\).

Let \(T:V\to W\) be linear. Let \(\cB=\{\bu_1,\ldots,\bu_n\}\) be a basis for \(V\) and \(\cB'=\{\bw_1,\ldots,\bw_m\}\) a basis for \(W\). Then \begin{equation*} [T]_{\cB',\cB}=\bigg[[T(\bu_1)]_{\cB'}\, \cdots\, [T(\bu_n)]_{\cB'}\bigg] \end{equation*}

Let \(T:U\to V\) and \(S:V\to W\) be linear. Let \(\cB,\cB',\cB''\) be bases for vector spaces \(U,V,W\) respectively. Then \([S\circ T]_{\cB'',\cB}=[S]_{\cB'',\cB'}\cdot [T]_{\cB',\cB}\).

Let \(\cB,\cB'\) be bases for a vector space \(V\). Then

1. For any \(\bv\) in \(V\), \([\bv]_{\cB'}=[I]_{\cB',\cB}\cdot [\bv]_\cB\).
2. \([I]_{\cB,\cB}=I_n\), where \(n=\dim V\).
3. \([I]_{\cB',\cB}\) is invertible.
4. \(([I]_{\cB',\cB})^{-1}=[I]_{\cB,\cB'}\).

Let \(T:V\to V\) be a linear operator. Let \(\cB,\cB'\) be bases for \(V\). Then \begin{equation*} [T]_{\cB'}=[I]_{\cB',\cB}\cdot [T]_{\cB}\cdot [I]_{\cB,\cB'} \end{equation*}

Let \(V\) be an inner product space. Then

1. \(\ip{\bv}{\bz}=0\) and \(\ip{\bz}{\bv}=0\), for every \(\bv\) in \(V\).
2. \(\ip{c_1\bv_1+\cdots +c_n\bv_n}{\bw}=c_1\ip{\bv_1}{\bw}+\cdots+c_n\ip{\bv_n}{\bw}\), for all scalars \(c_1,\ldots,c_n\) and vectors \(\bv_1,\ldots,\bv_n,\bw\).

Let \(V\) be an inner product space. Let \(\bu,\bv,\bw\) be in \(V\) and let \(c\) be a scalar. Then

1. \(\ip{\bu}{\bv+\bw}=\ip{\bu}{\bv}+\ip{\bu}{\bw}\).
2. \(\ip{\bu}{c\bv}=c\ip{\bu}{\bv}\).

Let \(V\) be an inner product space. Let \(\bv\) be in \(V\) and let \(c\) be a scalar. Then

1. \(\|c\bv\|=|c|\|\bv\|\).
2. \(\dfrac{\bv}{\|\bv\|}\) is a unit vector, if \(\bv\neq \bz\).

If \(S=\{\bu_1,\ldots,\bu_n\}\) is an orthogonal set of nonzero vectors in an inner product space, then \(S\) is linearly independent.

Let \(V\) be an inner product space, and let \(B=\{\bv_1,\ldots,\bv_n\}\) be an orthogonal basis for \(V\). Then for \(\bu\) in \(V\), \(\bu=c_1\bv_1+\cdots+c_n\bv_n\), where \begin{equation*} c_i=\frac{\ip{\bu}{\bv_i}}{\|\bv_i\|^2},\ \text{ for }i=1,\ldots,n. \end{equation*} If \(B\) is an orthonormal basis, then \begin{equation*} c_i=\ip{\bu}{\bv_i}\ \text{ for }i=1,\ldots,n. \end{equation*}

Let \(W\) be a subspace of an inner product space \(V\). Then

1. \(W^{\perp}\) is a subspace of \(V\); and
2. \(W\cap W^{\perp}=\{\bz\}\).

Let \(\bu,\bv\) be vectors in an inner product space \(V\) with \(\bv\neq \bz\). Let \({L=\sspan\{\bv\}}\), a one-dimensional subspace of \(V\). The we can uniquely write \({\bu=\by+\bz}\), with \(\by\) in \(L\) and \(\bz\) in \(L^{\perp}\). Explicitly, \begin{equation*} \by = \frac{\ip{\bu}{\bv}}{\|\bv\|^2}\bv,\ \text{ and }\ \bz=\bu-\by. \end{equation*} The vector \(\by\) is called the orthogonal projection of \(\bu\) onto \(L\) and denoted by \(\proj_L\bu\) or \(\proj_{\bv}\bu\).

Let \(\bu\) be a nonzero vector in an inner product space \(V\), and let \(W\) be a finite dimensional subspace of \(V\). Then we can uniquely write \(\bu=\by+\bz\), with \(\by\) in \(W\) and \(\bz\) in \(W^{\perp}\). The vector \(\by\) is called the orthogonal projection of \(\bu\) onto \(W\) and denoted by \(\proj_W\bu\), and \(\bz\) is called the component of \(\bu\) orthogonal to \(W\).

Let \(A\) be a square matrix.

1. If two rows of \(A\) are interchanged to produce a matrix \(B\), then \({\det B=-\det A}\).
2. If one row of \(A\) is multiplied by a constant \(k\) to produce \(B\), then \({\det B=k\det A}\).
3. If a multiple of one row of \(A\) is added to another row to produce \(B\), then \({\det B=\det A}\).

Let \(A\) be a square matrix. Then \(A\) is invertible if and only if \({\det A\neq 0}\).

Let \(A\) be an \(n\times n\) matrix. Then \(\lambda\) is an eigenvalue of \(A\) if and only if \(\det(\lambda I_n-A)=0\).

Let \(A\) be an \(n\times n\) matrix, and let \(\lambda\) be an eigenvalue of \(A\). Then \(\bx\) is an eigenvector of \(A\) corresponding to \(\lambda\) if and only if \(\bx\neq \bz\) and \(\bx\) is in \({\nul(\lambda I_n-A)}\).

Let \(A\) be an \(n\times n\) matrix, and let \(T:\bR^n\to \bR^n\) be the matrix transformation \(T(\bx)=A\bx\). Suppose that \(B=\{\bv_1,\ldots,\bv_n\}\) is a basis for \(\bR^n\) consisting of eigenvectors for \(A\) (i.e., an eigenbasis for \(A\)). Suppose that the eigenvalues of \(\bv_1,\ldots,\bv_n\) are \(\lambda_1,\ldots,\lambda_n\). Then \([T]_B\) is the following diagonal matrix: \begin{equation*} [T]_B= \begin{bmatrix} \lambda_1 & & &\\ & \lambda_2 & & \\ & & \ddots & \\ & & & \lambda_n \end{bmatrix} \end{equation*} If \(B'\) is the standard basis for \(\bR^n\), then by the change of basis theorem, \( [T]_{B'}=[I]_{B',B}[T]_B[I]_{B,B'} \). This is often written \(A=PDP^{-1}\).

Let \(A\) be an \(n\times n\) matrix. If \(\lambda_1,\ldots,\lambda_k\) are distinct eigenvalues of \(A\), and if \(\bv_1,\ldots,\bv_k\) are corresponding eigenvectors, then \(\{\bv_1,\ldots,\bv_k\}\) is linearly independent.

Let \(A\) be an \(n\times n\) matrix, and let \(\lambda_1,\ldots,\lambda_k\) be the distinct eigenvalues of \(A\).

1. The geometric multiplicity of any eigenvalue is less than or equal to its algebraic multiplicity.
2. \(A\) is diagonalizable if and only if the geometric multiplicity of each eigenvalue is equal to its algebraic multplicity.