Linear algebra formulas

The order reverses for any number of factors, e.g. $(ABC{)}^{-1}=C^{-1}{B}^{-1}{A}^{-1}$. The transpose and conjugate-transpose laws hold whenever $AB$ is defined, square or not; the inverse laws need every factor square and non-singular.

$A$ non-singular, $k\ne 0$ a scalar and $m$ a positive integer. A diagonal matrix with non-zero diagonal entries $d_1,\dots ,d_n$ has inverse $\mathrm{diag}(1/d_1,\dots ,1/d_n)$, and the inverse of a non-singular upper (lower) triangular matrix is again upper (lower) triangular.

The first identity holds for every square matrix; the inverse formula needs $|A|\ne 0$.

Swap the diagonal entries, change the sign of the off-diagonal entries, divide by the determinant. Valid only when $ad-bc\ne 0$.

$\mathrm{tr}(AB)=\mathrm{tr}(BA)$ holds whenever both products are defined, even for non-square $A$ and $B$. The trace is unchanged by cyclic shifts, $\mathrm{tr}(ABC)=\mathrm{tr}(BCA)=\mathrm{tr}(CAB)$, but not by other reorderings: in general $\mathrm{tr}(ABC)\ne \mathrm{tr}(ACB)$. In general $\mathrm{tr}(AB)\ne \mathrm{tr}A\mathrm{tr}B$.

$A$ and $B$ are $n\times n$ and $k$ is a scalar. $|A^m|=|A{|}^m$ holds for every positive integer $m$ (for every integer $m$ when $A$ is non-singular), and $|A^{-1}|=1/|A|$ needs $|A|\ne 0$. $|AB|=|BA|$ even when $AB\ne BA$.

For $n\times n$ matrices with $n\ge 2$; all four hold for singular matrices too. The first follows from taking determinants of $A\mathrm{adj}A=|A|I$, which gives $|A||\mathrm{adj}A|=|A{|}^n$. Consequently $|\mathrm{adj}(\mathrm{adj}A)|=|A{|}^{(n-1{)}^2}$. Also $\mathrm{adj}(A^T)=(\mathrm{adj}A{)}^T$, and for non-singular $A$, $\mathrm{adj}(A^{-1})=(\mathrm{adj}A{)}^{-1}=A/|A|$.

For real vectors in ${\mathbb{R}}^n$. For complex vectors use $u^{\ast }v$ and $\Vert u\Vert =\sqrt{u^{\ast }u}$. A set of vectors is orthonormal if they are mutually orthogonal and each has length $1$. In an orthonormal basis $e_1,\dots ,e_n$ the coordinates of any $v$ are simply $c_i=e_i^{T}v$.

$U$ and $W$ are subspaces of a finite-dimensional space and $U+W$ is the set of all sums $u+w$. Two distinct planes through the origin in ${\mathbb{R}}^3$ meet in a line: $3=2+2-1$.

For an $m\times n$ matrix. A square $n\times n$ matrix has rank $n$ exactly when $|A|\ne 0$.

$A$ is $m\times n$ and $B$ is $n\times p$, so $n$ is the shared inner dimension. Multiplying by a non-singular matrix does not change the rank: $\mathrm{rank}(PA)=\mathrm{rank}(AQ)=\mathrm{rank}A$ for invertible $P$ and $Q$.

$n$ is the number of columns of $A$ (the number of unknowns in $AX=0$), and $\mathrm{nullity}A$ is the dimension of the solution space of $AX=0$.

When consistent, the solution is unique if this common rank equals the number of unknowns $n$, and has $n-r$ free parameters if the common rank is $r<n$.

$A$ is $n\times n$. The number of linearly independent solutions is $n-\mathrm{rank}A$.

${\Delta }_i$ is the determinant of $A$ with its $i$-th column replaced by $B$. Applies only to a square system with $\Delta \ne 0$. If $\Delta =0$ and some ${\Delta }_i\ne 0$, the system is inconsistent.

For $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$: ${\lambda }^2-(a+d)\lambda +(ad-bc)=0$, so $\lambda =\frac{(a+d)\pm \sqrt{(a-d{)}^2+4bc}}{2}$. The eigenvalues are real whenever $bc\ge 0$, in particular whenever $A$ is real symmetric.

This is $|A-\lambda I|=0$ multiplied through by $-1$. $C_{11}+C_{22}+C_{33}$ is the sum of the three principal $2\times 2$ minors (the cofactors of the diagonal entries); it also equals ${\lambda }_1{\lambda }_2+{\lambda }_2{\lambda }_3+{\lambda }_3{\lambda }_1$.

Each eigenvalue is counted as often as its algebraic multiplicity, complex ones included. More generally $\mathrm{tr}(A^m)={\lambda }_1^m+{\lambda }_2^m+\cdots +{\lambda }_n^m$ for every positive integer $m$; for instance $\mathrm{tr}(A^2)$ is the sum of the squares of the eigenvalues. A quick check on any eigenvalue calculation.

An eigenvalue with AM $=1$ (a simple root) always has GM $=1$. A repeated eigenvalue can have any GM from $1$ up to its AM.

$f(D)=\mathrm{diag}(f({\lambda }_1),\dots ,f({\lambda }_n))$ acts entry by entry on the diagonal. $f$ is a polynomial, or a convergent power series such as the exponential: $e^{At}=P\mathrm{diag}(e^{{\lambda }_{1}t},\dots ,e^{{\lambda }_{n}t})P^{-1}$. For a real symmetric $A$, take orthonormal eigenvectors as the columns of $P$ (normalise each one and, inside the eigenspace of a repeated eigenvalue, also make them orthogonal); then $P^{-1}=P^T$. Normalising alone is not enough when an eigenvalue repeats: $(1,-1,0{)}^T$ and $(1,0,-1{)}^T$ in the Section 5 example are not orthogonal.

Using $|A-\lambda I|$ instead of $\det (\lambda I-A)$ only multiplies the polynomial by $(-1{)}^n$, which does not change the conclusion.

Multiply the Cayley–Hamilton identity by $A^{-1}$. Needs $c_0\ne 0$, that is, $|A|\ne 0$.

For any $2\times 2$ matrix; the inverse needs $|A|\ne 0$.

Here $p$ is the characteristic polynomial (degree $n$), and the result follows because $p(A)=O$. With distinct eigenvalues, find the $n$ coefficients of $r$ from $r({\lambda }_i)={\lambda }_i^m$. For an eigenvalue of algebraic multiplicity $k$, also match the first $k-1$ derivatives; for a double root this is $r^{\prime }({\lambda }_i)=m{\lambda }_i^{m-1}$.

Under a rotation of axes, with $a_{ij}$ the cosine of the angle between the new axis $x_i^{\prime }$ and the old axis $x_j$. The matrix $[a_{ij}]$ is orthogonal, $a_{ik}{a}_{jk}={\delta }_{ij}$. A third-order tensor picks up three factors: $T_{ijk}^{\prime }=a_{ip}{a}_{jq}{a}_{kr}{T}_{pqr}$.