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Matrix Operations and Algebra · Tutorial 83 of 1000

Powers Of Triangular Matrices

Use matrix multiplication and induction to establish the triangular form and diagonal entries of powers, then check the results in exact examples.

Beginner 9 min read

What You'll Learn

  • Define upper and lower triangular matrices by their entries.
  • Prove that products of triangular matrices of the same type retain that type.
  • Show that the diagonal of a product of triangular matrices is the product of their diagonals.
  • Apply induction to determine the diagonal entries of every matrix power.
  • Compute and verify powers of upper and lower triangular matrices.
  • Avoid confusing matrix powers with entry-by-entry powers.

Triangular Form Persists Under Powers

Multiplying a matrix by itself repeatedly can create many terms in each entry. For a triangular matrix, however, the zero pattern restricts which terms can contribute. As a result, every positive integer power keeps the same triangular form, and its diagonal entries are particularly simple: each is the corresponding diagonal entry of the original matrix raised to that power.

Throughout this tutorial, matrices have real entries. An \(n\times n\) matrix \(\mathbf{A}=[a_{ij}]\) is upper triangular if \(a_{ij}=0\) whenever \(i>j\). It is lower triangular if \(a_{ij}=0\) whenever \(i<j\). Thus, an upper triangular matrix has zeros below its main diagonal, and a lower triangular matrix has zeros above it. A diagonal matrix is both upper and lower triangular.

For a square matrix \(\mathbf{A}\), define \(\mathbf{A}^1=\mathbf{A}\), and for each integer \(k\geq 1\), define \(\mathbf{A}^{k+1}=\mathbf{A}^k\mathbf{A}\). Also set \(\mathbf{A}^0=\mathbf{I}_n\), the \(n\times n\) identity matrix. The multiplication order in this recursive definition is unambiguous because powers of the same matrix can be associated using associativity of matrix multiplication.

Theorem: Let \(\mathbf{A}\in\mathbb{R}^{n\times n}\). If \(\mathbf{A}\) is upper triangular, then \(\mathbf{A}^k\) is upper triangular for every integer \(k\geq 0\), and its \(i\)th diagonal entry is \(a_{ii}^k\) for \(k\geq 1\). The same statements hold when \(\mathbf{A}\) is lower triangular. For \(k=0\), every diagonal entry of \(\mathbf{A}^0\) is \(1\).

The theorem says nothing that would make the off-diagonal entries powers of the original entries. Those entries come from sums of products, as the examples will show. The key fact is that, in a product of two triangular matrices of the same type, the zero pattern is preserved and each diagonal entry is just a product of the corresponding diagonal entries.

Why Triangular Products Have the Required Pattern

For any two \(n\times n\) matrices \(\mathbf{A}\) and \(\mathbf{B}\), the entry in row \(i\), column \(j\) of their product is \((\mathbf{A}\mathbf{B})_{ij}=\sum_{k=1}^{n}a_{ik}b_{kj}\). Each term uses an entry from row \(i\) of \(\mathbf{A}\) and an entry from column \(j\) of \(\mathbf{B}\). Triangular zeros can make all terms vanish when the entry lies on the wrong side of the diagonal.

Proof. First suppose \(\mathbf{A}\) and \(\mathbf{B}\) are upper triangular. If \(i>j\), a nonzero term \(a_{ik}b_{kj}\) would require \(i\leq k\), because \(\mathbf{A}\) is upper triangular, and \(k\leq j\), because \(\mathbf{B}\) is upper triangular. These inequalities would imply \(i\leq j\), contradicting \(i>j\). Therefore every term in \((\mathbf{A}\mathbf{B})_{ij}\) is zero whenever \(i>j\), so \(\mathbf{A}\mathbf{B}\) is upper triangular.

Now suppose \(\mathbf{A}\) and \(\mathbf{B}\) are lower triangular. If \(i<j\), a nonzero term \(a_{ik}b_{kj}\) would require \(k\leq i\) and \(k\geq j\), respectively. Together these imply \(j\leq i\), contradicting \(i<j\). Thus every entry above the diagonal of \(\mathbf{A}\mathbf{B}\) is zero, so the product is lower triangular.

For either type, consider a diagonal entry \((\mathbf{A}\mathbf{B})_{ii}=\sum_{k=1}^{n}a_{ik}b_{ki}\). In the upper triangular case, a possibly nonzero term requires \(i\leq k\) and \(k\leq i\), hence \(k=i\). In the lower triangular case, it requires \(i\geq k\) and \(k\geq i\), again giving \(k=i\). In both cases, \((\mathbf{A}\mathbf{B})_{ii}=a_{ii}b_{ii}\). This proves the product and diagonal claims used below. \(\square\)

To obtain the theorem about powers, use induction on \(k\). The initial positive power is \(\mathbf{A}^1=\mathbf{A}\), which has the assumed triangular form and diagonal entries \(a_{ii}\). If \(\mathbf{A}^k\) has the same triangular form, then \(\mathbf{A}^{k+1}=\mathbf{A}^k\mathbf{A}\) has that form by the product result. Its \(i\)th diagonal entry is \((\mathbf{A}^k)_{ii}a_{ii}=a_{ii}^k a_{ii}=a_{ii}^{k+1}\). Induction proves the claims for every \(k\geq1\). At \(k=0\), \(\mathbf{A}^0=\mathbf{I}_n\), which is triangular and has diagonal entries \(1\). This completes the proof of the theorem.

Computing Powers of Upper Triangular Matrices

Worked Example: A Two-by-Two Upper Triangular Matrix

Let \(\mathbf{A}=\begin{bmatrix}2&1\\0&3\end{bmatrix}\). Compute \(\mathbf{A}^2\) and \(\mathbf{A}^3\), and check their diagonal entries.

Multiply \(\mathbf{A}\) by itself, computing each entry as a row-column product:

$$ \mathbf{A}^2= \begin{bmatrix}2&1\\0&3\end{bmatrix} \begin{bmatrix}2&1\\0&3\end{bmatrix} = \begin{bmatrix} 2\cdot2+1\cdot0 & 2\cdot1+1\cdot3\\ 0\cdot2+3\cdot0 & 0\cdot1+3\cdot3 \end{bmatrix} = \begin{bmatrix}4&5\\0&9\end{bmatrix}. $$

Now multiply this result by \(\mathbf{A}\):

$$ \mathbf{A}^3=\mathbf{A}^2\mathbf{A} = \begin{bmatrix}4&5\\0&9\end{bmatrix} \begin{bmatrix}2&1\\0&3\end{bmatrix} = \begin{bmatrix} 4\cdot2+5\cdot0 & 4\cdot1+5\cdot3\\ 0\cdot2+9\cdot0 & 0\cdot1+9\cdot3 \end{bmatrix} = \begin{bmatrix}8&19\\0&27\end{bmatrix}. $$

Check: The diagonal entries of \(\mathbf{A}^2\) are \(2^2=4\) and \(3^2=9\); those of \(\mathbf{A}^3\) are \(2^3=8\) and \(3^3=27\). Both matrices remain upper triangular. The off-diagonal entries, \(5\) and \(19\), are not obtained by simply squaring or cubing the original entry \(1\).

This example illustrates the difference between the diagonal rule and the full matrix power. The diagonal rule predicts two entries directly, but it does not by itself compute the other entries. To find those, perform the matrix multiplication or use an additional result specific to the matrix.

Worked Example: A Three-by-Three Upper Triangular Matrix

Let \(\mathbf{B}=\begin{bmatrix}2&1&1\\0&2&1\\0&0&0\end{bmatrix}\). Compute \(\mathbf{B}^2\) and \(\mathbf{B}^3\). Keep the full row-column products visible, since the upper-right entries contain contributions from more than one multiplication path.

First compute \(\mathbf{B}\mathbf{B}\):

$$ \mathbf{B}^2= \begin{bmatrix} 2\cdot2+1\cdot0+1\cdot0 & 2\cdot1+1\cdot2+1\cdot0 & 2\cdot1+1\cdot1+1\cdot0\\ 0\cdot2+2\cdot0+1\cdot0 & 0\cdot1+2\cdot2+1\cdot0 & 0\cdot1+2\cdot1+1\cdot0\\ 0&0&0 \end{bmatrix} = \begin{bmatrix}4&4&3\\0&4&2\\0&0&0\end{bmatrix}. $$

Next multiply \(\mathbf{B}^2\) by \(\mathbf{B}\):

$$ \mathbf{B}^3= \begin{bmatrix}4&4&3\\0&4&2\\0&0&0\end{bmatrix} \begin{bmatrix}2&1&1\\0&2&1\\0&0&0\end{bmatrix} = \begin{bmatrix} 8&4+8&4+4\\ 0&8&4\\ 0&0&0 \end{bmatrix} = \begin{bmatrix}8&12&8\\0&8&4\\0&0&0\end{bmatrix}. $$

Check: The diagonal of \(\mathbf{B}\) is \(2,2,0\), so the diagonal of \(\mathbf{B}^2\) must be \(4,4,0\), and the diagonal of \(\mathbf{B}^3\) must be \(8,8,0\); both computed results agree. The potentially easy-to-miscompute entry in row 2, column 3 of \(\mathbf{B}^3\) is the row-column product \((0,4,2)\cdot(1,1,0)=0\cdot1+4\cdot1+2\cdot0=4\).

Computing Powers of Lower Triangular Matrices

The lower triangular case follows the same product rule, but the zeros lie above the diagonal instead. A useful check is to inspect the entries above the diagonal after multiplying; they must remain zero. The diagonal still comes from multiplying corresponding diagonal entries.

Worked Example: Squaring a Lower Triangular Matrix

Let \(\mathbf{C}=\begin{bmatrix}1&0&0\\2&2&0\\1&-1&3\end{bmatrix}\). Compute \(\mathbf{C}^2\) by row-column products and verify its triangular form and diagonal.

Multiplication gives

$$ \mathbf{C}^2= \begin{bmatrix} 1\cdot1+0\cdot2+0\cdot1 & 1\cdot0+0\cdot2+0\cdot(-1) & 1\cdot0+0\cdot0+0\cdot3\\ 2\cdot1+2\cdot2+0\cdot1 & 2\cdot0+2\cdot2+0\cdot(-1) & 2\cdot0+2\cdot0+0\cdot3\\ 1\cdot1+(-1)\cdot2+3\cdot1 & 1\cdot0+(-1)\cdot2+3\cdot(-1) & 1\cdot0+(-1)\cdot0+3\cdot3 \end{bmatrix} = \begin{bmatrix}1&0&0\\6&4&0\\2&-5&9\end{bmatrix}. $$

Check: The entries above the diagonal are zero, as required for a lower triangular matrix. The diagonal entries are \(1^2=1\), \(2^2=4\), and \(3^2=9\), as the theorem predicts. The lower off-diagonal entries need not be the squares of the corresponding entries of \(\mathbf{C}\): for instance, the entry in row 3, column 2 is \(-5\), not \((-1)^2=1\).

Common Mistake: Raising Entries Instead of the Matrix

A matrix power means repeated matrix multiplication, not raising each entry to a power separately. For example, squaring \(\begin{bmatrix}2&1\\0&3\end{bmatrix}\) entry by entry would incorrectly give \(\begin{bmatrix}4&1\\0&9\end{bmatrix}\). The actual matrix square is \(\begin{bmatrix}4&5\\0&9\end{bmatrix}\), because its upper-right entry is \(2\cdot1+1\cdot3=5\). The diagonal entries happen to be the squares of the original diagonal entries, but this is a consequence of triangular matrix multiplication, not a rule for all entries.

Another useful safeguard is to distinguish the guaranteed conclusions from the calculation. Triangular form and diagonal powers follow from the theorem, so they can be checked without computing every entry. The off-diagonal entries still require the usual multiplication. In an upper triangular product, entries below the diagonal vanish; in a lower triangular product, entries above it vanish. If a computed power violates that pattern, there is an arithmetic or matrix-layout error.

These facts make powers of triangular matrices easier to organize: the diagonal can be predicted at once, and the zero pattern reduces the number of products that need to be calculated. They also provide a fast consistency check on longer computations. They do not determine every off-diagonal entry, so the product itself must still be carried out carefully whenever those entries are needed.

Check Your Understanding

Use the definitions and product rule above. When you compute a power, check both its zero pattern and its diagonal entries.

  1. For \(\begin{bmatrix}1&2\\0&-2\end{bmatrix}\), compute the square and verify the diagonal entries.
  2. State the diagonal entries of \(\mathbf{A}^4\) if an upper triangular matrix \(\mathbf{A}\) has diagonal entries \(3,0,-1\).
  3. For \(\begin{bmatrix}2&0\\1&3\end{bmatrix}\), compute the square and verify that it remains lower triangular.
  4. In a product of two upper triangular matrices, explain why every term in entry \((i,j)\) vanishes when \(i>j\).
  5. Identify the error in claiming that the square of \(\begin{bmatrix}1&4\\0&2\end{bmatrix}\) is \(\begin{bmatrix}1&16\\0&4\end{bmatrix}\).

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