Impose Every Matrix Condition at Once
A matrix may be required to satisfy several structural conditions simultaneously. It might be symmetric and upper triangular, or symmetric with specified entries equal to zero. The intersection consists precisely of the matrices that satisfy all the conditions. A reliable method is to write a general matrix in the ambient space, impose each condition on its entries, and then check that the resulting description still satisfies every condition.
Throughout, the scalar field is \(\mathbb{R}\), and the ambient space for the examples is \(\mathbb{R}^{3\times 3}\), the vector space of real \(3\times3\) matrices. The same reasoning works for other matrix sizes. A subspace of an ambient vector space is a subset that contains the zero vector and is closed under addition and scalar multiplication. Equivalently, it contains every linear combination of any two of its elements.
For an entry restriction such as \(a_{12}=0\), the indices specify row 1 and column 2. Several entry restrictions can be imposed together, for example \(a_{12}=0\) and \(a_{13}=a_{23}\). These equations constrain the entries; solving them reveals the matrices in the intersection. When the equations are homogeneous linear constraints, the resulting sets are subspaces, as we verify below.
Why Intersections Remain Subspaces
Proof. Since \(S\) and \(T\) are subspaces, both contain the zero vector of \(V\), so \(\mathbf{0}\in S\cap T\). Now take any \(\mathbf{x},\mathbf{y}\in S\cap T\) and scalars \(a,b\in\mathbb{R}\). Membership in the intersection means \(\mathbf{x},\mathbf{y}\in S\) and \(\mathbf{x},\mathbf{y}\in T\). Since \(S\) is a subspace, \(a\mathbf{x}+b\mathbf{y}\in S\); since \(T\) is a subspace, \(a\mathbf{x}+b\mathbf{y}\in T\). Hence \(a\mathbf{x}+b\mathbf{y}\in S\cap T\). The intersection contains zero and is closed under linear combinations, so it is a subspace. \(\square\)
This theorem guarantees closure once the sets being intersected are known to be subspaces. For concrete matrix problems, it is still useful to check the resulting description directly: doing so confirms that the entry restrictions were translated correctly and shows how the matrices can be built.
Symmetry Together with Upper Triangular Form
Worked Example: Symmetric and Upper Triangular Matrices
Find all matrices in \(\mathbb{R}^{3\times3}\) that are both symmetric and upper triangular.
Start with an arbitrary matrix. Upper triangular form sets the three entries below the diagonal to zero:
For symmetry, corresponding entries across the diagonal must agree. In particular, \(a_{12}=a_{21}\), so \(b=0\); \(a_{13}=a_{31}\), so \(c=0\); and \(a_{23}=a_{32}\), so \(e=0\). The diagonal entries impose no further conditions. Therefore the intersection is
To check that this description is exact, each displayed matrix is symmetric because its transpose is unchanged, and it is upper triangular because every entry below the diagonal is zero. Conversely, the symmetry equations just derived force every off-diagonal entry to be zero in any matrix in the intersection.
For a particular member, take \(a=2\), \(d=-1\), and \(f=3\). Then \(\mathbf{A}=\begin{bmatrix}2&0&0\\0&-1&0\\0&0&3\end{bmatrix}\), and \(\mathbf{A}^{\top}=\begin{bmatrix}2&0&0\\0&-1&0\\0&0&3\end{bmatrix}=\mathbf{A}\); its entries below the diagonal are all zero.
The description also verifies closure directly. If \(\mathbf{D}_1=\begin{bmatrix}a_1&0&0\\0&d_1&0\\0&0&f_1\end{bmatrix}\) and \(\mathbf{D}_2=\begin{bmatrix}a_2&0&0\\0&d_2&0\\0&0&f_2\end{bmatrix}\), then for \(r,s\in\mathbb{R}\),
This has the same form, including when \(r=s=0\), so the intersection is a subspace.
The important feature of this example is that symmetry relates entries in pairs, whereas upper triangular form sets some of those entries to zero. Once a lower-triangular entry is zero, symmetry forces its matching upper-triangular entry to be zero as well. Checking only the diagonal entries or only the triangular condition would miss this consequence.
Symmetry with Entry Restrictions
Worked Example: A Symmetric Matrix with Two Restrictions
Find the symmetric \(3\times3\) matrices satisfying \(a_{12}=0\) and \(a_{13}=a_{23}\).
Write a general symmetric matrix using one variable for each pair of matching off-diagonal entries:
The first entry restriction gives \(b=0\). The second gives \(c=e\); call their common value \(t\). The matrices in the intersection are therefore exactly
For verification, the transpose of this matrix is \(\begin{bmatrix}a&0&t\\0&d&t\\t&t&f\end{bmatrix}\), so it is symmetric. Its \((1,2)\) entry is zero, and its \((1,3)\) and \((2,3)\) entries both equal \(t\), as required. The derivation also proves the converse: every symmetric matrix satisfying the two restrictions must have this form.
For example, setting \(a=2\), \(d=4\), \(t=-1\), and \(f=3\) gives
Here \(\mathbf{A}^{\top}=\mathbf{A}\), \(a_{12}=0\), and \(a_{13}=-1=a_{23}\). To check closure in the general form, let \(\mathbf{A}_1\) and \(\mathbf{A}_2\) have parameters \((a_1,d_1,t_1,f_1)\) and \((a_2,d_2,t_2,f_2)\). For real \(r,s\), their linear combination is
It remains symmetric, its \((1,2)\) entry is zero, and its \((1,3)\) and \((2,3)\) entries are equal. Thus the simultaneous restrictions are preserved.
In this calculation, symmetry was used first to represent the matrix with fewer independent entry variables, and then the additional equations were imposed. One could instead begin with nine entries and impose all the equations, but using the symmetry relations at the outset avoids repeating equivalent conditions.
Triangular Form with Entry Restrictions
Worked Example: An Upper Triangular Matrix with Equal Diagonal Entries
Find all upper triangular \(3\times3\) matrices satisfying \(a_{11}=a_{22}\) and \(a_{13}=0\).
Begin with the general upper triangular matrix, so its entries below the diagonal are already zero:
The condition \(a_{11}=a_{22}\) says \(a=d\), and \(a_{13}=0\) says \(c=0\). Substitution gives the complete description
Every matrix of this form is upper triangular, has equal first and second diagonal entries, and has a zero \((1,3)\) entry. Conversely, the two added equations force exactly the substitutions made above, so no other matrices belong to the intersection.
For instance, choose \(a=2\), \(b=-1\), \(e=3\), and \(f=4\). The resulting matrix is \(\begin{bmatrix}2&-1&0\\0&2&3\\0&0&4\end{bmatrix}\). Its entries below the diagonal are \(0\); its first and second diagonal entries are both \(2\); and its \((1,3)\) entry is \(0\).
Finally, for two matrices of the displayed general form, with parameters \((a_1,b_1,e_1,f_1)\) and \((a_2,b_2,e_2,f_2)\), a linear combination with scalars \(r,s\) is
This is still upper triangular, its first two diagonal entries still agree, and its \((1,3)\) entry is still zero. This confirms closure directly.
A Reliable Procedure and a Common Mistake
For a matrix-intersection problem, the following sequence keeps the conditions separate until they have all been applied:
State the matrix size and scalar field, then write a general matrix with an entry in every position.
Use symmetry, triangular form, or another specified structure to write the required entry equations.
Substitute the equations into the general form and retain only the free real parameters.
Check that the proposed form satisfies every original condition, then confirm the zero matrix and closure under linear combinations.
Common mistake: treating an intersection as an “or” condition. For example, consider the \(2\times2\) intersection of symmetric matrices and upper triangular matrices. The matrix \(\begin{bmatrix}1&2\\0&3\end{bmatrix}\) is upper triangular, but it is not symmetric because its transpose is \(\begin{bmatrix}1&0\\2&3\end{bmatrix}\), which differs from the original matrix. It is therefore not in the intersection. Applying symmetry as well forces the off-diagonal entries to be zero, leaving only diagonal matrices \(\begin{bmatrix}a&0\\0&d\end{bmatrix}\). The test is always whether every defining condition holds at once.
The subspace theorem explains why an intersection of subspaces remains a subspace, but it does not replace solving the entry equations. The equations describe which matrices belong to the intersection; the closure argument verifies that linear combinations of those matrices still belong. Keeping these two tasks distinct makes it easier to detect both algebra mistakes and incorrect subspace claims.
Check Your Understanding
For each problem, write a general matrix, impose all the conditions, and check the resulting form.
- Describe the \(2\times2\) matrices that are both symmetric and upper triangular.
- Find all symmetric \(3\times3\) matrices satisfying \(a_{13}=0\) and \(a_{12}=a_{23}\).
- Find all upper triangular \(3\times3\) matrices satisfying \(a_{11}=a_{33}\) and \(a_{12}=0\).
- Explain why a matrix that is symmetric but not upper triangular cannot belong to the intersection of the symmetric and upper triangular matrix subspaces.
- Use the intersection theorem to explain why the common matrices in any two matrix subspaces form a subspace.
