Two Special Matrices and Their Effects
Some matrices have a simple, predictable effect under multiplication: an identity matrix leaves a compatible matrix unchanged, while a zero matrix makes the product zero. These statements are simple to use, but the sizes matter. A rectangular matrix has different numbers of rows and columns, so the identity matrix that can multiply it on the left may not be the one that can multiply it on the right. Similarly, the dimensions of a zero product are determined by the product, not by the zero factor alone.
Throughout, matrices have real entries and positive integer dimensions. Recall that if \(\mathbf{A}\) has \(m\) rows and \(n\) columns, then its entry in row \(i\), column \(j\) is \(a_{ij}\), and we write \(\mathbf{A}\in\mathbb{R}^{m\times n}\). For a product \(\mathbf{A}\mathbf{B}\) to be defined, the number of columns of \(\mathbf{A}\) must equal the number of rows of \(\mathbf{B}\). Its \((i,j)\)-entry is the sum of products across row \(i\) of \(\mathbf{A}\) and column \(j\) of \(\mathbf{B}\).
The identity matrix is always square, but a zero matrix need not be square. The dimensions in the notation \(\mathbf{O}_{r\times s}\) specify its shape, including when \(r\ne s\). If \(\mathbf{A}\in\mathbb{R}^{m\times n}\), then left multiplication by an identity matrix requires an \(m\times m\) identity, while right multiplication requires an \(n\times n\) identity. Thus the dimension-matched products are \(\mathbf{I}_m\mathbf{A}\) and \(\mathbf{A}\mathbf{I}_n\).
Why Identity Matrices Leave Products Unchanged
Proof. Write the entries of \(\mathbf{I}_m\) as \(\delta_{ik}\), where \(\delta_{ik}=1\) if \(i=k\) and \(\delta_{ik}=0\) otherwise. The \((i,j)\)-entry of \(\mathbf{I}_m\mathbf{A}\) is \(\sum_{k=1}^{m}\delta_{ik}a_{kj}\). Every term is zero except the term with \(k=i\), so this sum is \(a_{ij}\). This holds for every \(i\) and \(j\), hence \(\mathbf{I}_m\mathbf{A}=\mathbf{A}\). For the product on the right, the \((i,j)\)-entry of \(\mathbf{A}\mathbf{I}_n\) is \(\sum_{k=1}^{n}a_{ik}\delta_{kj}\). Every term is zero except the term with \(k=j\), so this sum is \(a_{ij}\). Therefore \(\mathbf{A}\mathbf{I}_n=\mathbf{A}\) as well. Both products have the same size as \(\mathbf{A}\), so equality of all their entries proves the stated matrix equalities. \(\square\)
The two products work for different dimensional reasons. On the left, \(\mathbf{I}_m\) has \(m\) columns to match the \(m\) rows of \(\mathbf{A}\); on the right, \(\mathbf{I}_n\) has \(n\) rows to match the \(n\) columns of \(\mathbf{A}\). The entry calculation shows what each multiplication does: the left identity selects the matching row of \(\mathbf{A}\), and the right identity selects the matching column. Neither multiplication changes any entry.
Worked Example: Multiplying a Rectangular Matrix by Identities
Let \(\mathbf{A}=\begin{bmatrix}2&-1&3\\0&4&5\end{bmatrix}\), so \(\mathbf{A}\) is \(2\times3\). Compute the products with the correctly sized identities on both sides.
For left multiplication, the identity must be \(\mathbf{I}_2\). The two rows are calculated as follows:
For right multiplication, the identity must be \(\mathbf{I}_3\). Each column of \(\mathbf{A}\) is selected unchanged:
Check: both products are \(2\times3\), as required by their dimensions, and every calculated entry matches the corresponding entry of \(\mathbf{A}\).
Why Multiplication by a Zero Matrix Gives Zero
Proof. The first product is defined because \(\mathbf{O}_{r\times m}\) has \(m\) columns, matching the \(m\) rows of \(\mathbf{A}\). Its \((i,j)\)-entry, for \(1\le i\le r\) and \(1\le j\le n\), is \(\sum_{k=1}^{m}0\cdot a_{kj}=0\), since every entry in row \(i\) of the zero matrix is zero. The output has \(r\) rows and \(n\) columns, so it is exactly \(\mathbf{O}_{r\times n}\). The second product is defined because the \(n\) columns of \(\mathbf{A}\) match the \(n\) rows of \(\mathbf{O}_{n\times s}\). Its \((i,j)\)-entry, for \(1\le i\le m\) and \(1\le j\le s\), is \(\sum_{k=1}^{n}a_{ik}\cdot0=0\). This output has \(m\) rows and \(s\) columns, so it is \(\mathbf{O}_{m\times s}\). Thus both products are zero matrices of the stated sizes. \(\square\)
The entries vanish because each sum in the product formula consists only of zero terms. The dimensions, however, still follow the ordinary matrix-product rule: the product takes its row count from the left factor and its column count from the right factor. This is why the resulting zero matrix may have a different size from the zero matrix used as a factor.
Worked Example: Zero Matrices on Both Sides
Use \(\mathbf{A}=\begin{bmatrix}1&-2&0\\3&1&4\end{bmatrix}\), which is \(2\times3\). Compute \(\mathbf{O}_{4\times2}\mathbf{A}\) and \(\mathbf{A}\mathbf{O}_{3\times2}\).
The first product is \((4\times2)(2\times3)\), so its size is \(4\times3\). Each of its entries is a sum of two zero products:
The second product is \((2\times3)(3\times2)\), so its size is \(2\times2\):
Check: the first result has four rows from its left factor and three columns from its right factor; the second has two rows and two columns. Every entry in each result is zero, as the entry computations show.
Combining the Rules in a Product
The identity and zero rules can be used successively, as long as each product is dimensionally compatible. An identity factor preserves the matrix at that stage; a zero factor makes the next product zero. The order and sizes of the factors must still be checked before applying either rule.
Worked Example: An Identity Followed by a Zero Matrix
Let \(\mathbf{B}=\begin{bmatrix}1&-2&0\\3&1&4\end{bmatrix}\). Evaluate \(\mathbf{I}_2\mathbf{B}\mathbf{O}_{3\times1}\), showing the sizes at each stage.
The first product is defined because \(\mathbf{I}_2\) is \(2\times2\) and \(\mathbf{B}\) is \(2\times3\). It equals \(\mathbf{B}\):
Now multiply the resulting \(2\times3\) matrix by the \(3\times1\) zero matrix. The output must be \(2\times1\):
Check: the first multiplication leaves \(\mathbf{B}\) unchanged, and the second has one zero entry in each output row because its only column is zero. The final matrix is a \(2\times1\) zero matrix.
Common Mistakes About Sizes
Using an identity with the wrong dimensions. For the \(2\times3\) matrix \(\mathbf{A}\) in the first example, writing \(\mathbf{I}_3\mathbf{A}\) does not give \(\mathbf{A}\). In fact, that product is not defined: \(\mathbf{I}_3\) has \(3\) columns, but \(\mathbf{A}\) has \(2\) rows, and \(3\ne2\). The correction is to use \(\mathbf{I}_2\mathbf{A}\) on the left and \(\mathbf{A}\mathbf{I}_3\) on the right.
Giving the zero product the size of its zero factor. It would be wrong to say that \(\mathbf{O}_{4\times2}\mathbf{A}\) is \(4\times2\) just because its left factor has that size. The product’s inner dimensions, \(2\) and \(2\), match, and the outer dimensions are \(4\) and \(3\); the result is therefore \(4\times3\). Likewise, \(\mathbf{A}\mathbf{O}_{3\times2}\) is \(2\times2\), not \(3\times2\).
| Product | Required factor sizes | Result |
|---|---|---|
| \(\mathbf{I}_m\mathbf{A}\) | \(\mathbf{A}\) is \(m\times n\) | \(\mathbf{A}\), of size \(m\times n\) |
| \(\mathbf{A}\mathbf{I}_n\) | \(\mathbf{A}\) is \(m\times n\) | \(\mathbf{A}\), of size \(m\times n\) |
| \(\mathbf{O}_{r\times m}\mathbf{A}\) | \(\mathbf{A}\) is \(m\times n\) | \(\mathbf{O}_{r\times n}\) |
| \(\mathbf{A}\mathbf{O}_{n\times s}\) | \(\mathbf{A}\) is \(m\times n\) | \(\mathbf{O}_{m\times s}\) |
These matrices are useful as reference points for checking calculations: an identity factor should preserve all entries, and a zero factor should make every entry of the product vanish. They do not remove the need to check conformability. Before calculating, identify the number of rows and columns of every factor; then verify that the inner dimensions match and record the outer dimensions as the size of the result.
Check Your Understanding
For each question, state the dimensions before using the multiplication rules.
- If \(\mathbf{C}\) is \(3\times2\), which identity matrix leaves it unchanged on the left, and which leaves it unchanged on the right?
- What is \(\mathbf{O}_{5\times3}\mathbf{C}\) when \(\mathbf{C}\) is \(3\times2\)? State both the entries and the dimensions.
- If \(\mathbf{D}\) is \(4\times1\), is \(\mathbf{I}_4\mathbf{D}\) defined? Is \(\mathbf{D}\mathbf{I}_4\) defined? Explain by comparing inner dimensions.
- For \(\mathbf{E}\) of size \(2\times4\), give a zero matrix that can multiply \(\mathbf{E}\) on the right and make the result \(2\times3\).
- Explain why \(\mathbf{O}_{2\times3}\mathbf{O}_{3\times1}\) is a \(2\times1\) zero matrix rather than a \(2\times3\) or \(3\times1\) matrix.
