Adjoint Properties

Question

If the determinant of a matrix is 0 and therefore the matrix is not invertible, does the adjoint exist? If the matrix inverse is singular and therefore does not exist, does the adjoint exist?

Answer 1

Yes, for every matrix adjoint exists. But if determinant = 0. Inverse doen't exit. This is because we are dividing each element of adj matrix by determinant. So if determinant is zero then division is not possible (more correctly, not defined). Hence, inverse doesn't exits but adjoint always exits.

Answer 2

For a matrix $A\in\mathbb{R}^{n\times n}$ the adjoint with respect to the inner product $(x,y)=x^{tr}\cdot y$ for column vectors $x,y\in\mathbb{R^n}$ and where$\cdot :\mathbb{R}^{1\times n}\times\mathbb{R}^{n\times 1}\rightarrow\mathbb{R}$ is the usual matrix multiplication is given by $A^{tr}$ (the transpose of $A$) as can be seen from: $$(Ax,y)=(Ax)^{tr}\cdot y=(x^{tr}\cdot A^{tr})\cdot y=x^{tr}\cdot(A^{tr}\cdot y)=(x,A^{tr}y)$$ for all $x,y\in\mathbb{R}^n$.

For a matrix $A\in\mathbb{C}^{n\times n}$ the adjoint with respect to the inner product $(x,y)=x^{tr}\cdot\overline{ y}$ for column vectors $x,y\in\mathbb{R^n}$ and where$\cdot :\mathbb{C}^{1\times n}\times\mathbb{C}^{n\times 1}\rightarrow\mathbb{C}$ is the usual matrix multiplication and $\overline{y}$ denotes the vector with the complex conjugate entries is given by $\overline{A^{tr}}$ (the complex conjugate of the transpose of $A$) as can be seen from: $$(Ax,y)=(Ax)^{tr}\cdot \overline{ y}=(x^{tr}\cdot A^{tr})\cdot \overline{y}=x^{tr}\cdot\overline{(\overline{A^{tr}}\cdot y)}=(x,\overline{A^{tr}}y)$$ for all $x,y\in\mathbb{C}^n$. Clearly these matrices exist for every matrix, invertible or not. So the adjoint matrix always exists.

More generally for two Hilbert spaces $\mathcal{H_1}$ and $\mathcal{H_2}$ and a bounded linear operator $A:\mathcal{H_1}\rightarrow\mathcal{H}_2$ one obtains a bounded linear functional $x\mapsto (Ax,y)_{\mathcal{H_2}}$ on $\mathcal{H}_1$ for every $y\in\mathcal{H}_2$. By Riesz´srepresentation theorem there exists a unique element $x_y\in \mathcal{H}_1$ such that $$(Ax,y)_{\mathcal{H}_2}=(x,x_y)_{\mathcal{H}_1}$$ for all $x\in\mathcal{H}_1$. One defines $A^*:\mathcal{H_2\rightarrow\mathcal{H}_1}$ by $A^*y=x_y$ which turns out to be a bounded linear operator that constitutes the adjoint of $A$.