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Inner Products and Norms · Tutorial 609 of 1000

Frobenius Norms Of Matrices

Learn to calculate Frobenius norms from entries and interpret how sums, scalar multiples, and transposes affect matrix size.

Intermediate 9 min read

What You'll Learn

  • Calculate the Frobenius norm of a real or complex matrix from its entries.
  • Compare matrix sizes using exact Frobenius norm calculations.
  • Find the norm of a scalar multiple and identify the role of the scalar's absolute value.
  • Verify that transposing a matrix preserves its Frobenius norm.
  • Expand the squared norm of a sum and interpret its inner-product term.

Measuring the Size of a Matrix

A matrix has many entries, so comparing two matrices entry by entry can be awkward. The Frobenius norm packages the sizes of all the entries into one nonnegative number: square their absolute values, add the results, and take the square root. It is a useful way to quantify the overall size of a matrix, and its connection to the Frobenius inner product lets us understand how that size changes under addition.

For a real matrix, the absolute value of an entry is just its ordinary absolute value. For a complex matrix, the absolute value is the complex modulus. If \(\mathbf{A}\) and \(\mathbf{B}\) have the same dimensions, their sum is defined entry by entry. We can compare their Frobenius norms as numbers; a larger norm means a larger Frobenius size, not that every entry of one matrix is larger than the corresponding entry of the other.

Definition: For \(\mathbf{A}=(a_{ij})\in\mathbb{C}^{m\times n}\), the Frobenius norm is \(\|\mathbf{A}\|_F=\left(\sum_{i=1}^{m}\sum_{j=1}^{n}|a_{ij}|^2\right)^{1/2}\). For a real matrix, this reduces to \(\|\mathbf{A}\|_F=\left(\sum_{i=1}^{m}\sum_{j=1}^{n}a_{ij}^2\right)^{1/2}\). The subscript \(F\) identifies this particular matrix norm.

This formula comes from the Frobenius inner product: with the convention that the inner product is linear in its first slot, \(\langle\mathbf{A},\mathbf{B}\rangle=\sum_{i=1}^{m}\sum_{j=1}^{n}a_{ij}\overline{b_{ij}}\). Thus \(\|\mathbf{A}\|_F=\sqrt{\langle\mathbf{A},\mathbf{A}\rangle}\). In particular, \(\|\mathbf{A}\|_F\geq0\), and \(\|\mathbf{A}\|_F=0\) exactly when every entry is zero, that is, when \(\mathbf{A}=\mathbf{O}\). The Frobenius norm is the norm induced by this inner product.

Calculating and Comparing Norms

Worked Example: Comparing Two Matrices

Calculate and compare the Frobenius norms of \(\mathbf{A}=\begin{bmatrix}1&-2\\0&2\end{bmatrix}\) and \(\mathbf{B}=\begin{bmatrix}2&0\\-1&1\end{bmatrix}\).

For \(\mathbf{A}\), square each entry and add:

$$ \|\mathbf{A}\|_F =\sqrt{1^2+(-2)^2+0^2+2^2} =\sqrt{1+4+0+4} =\sqrt{9} =3. $$

For \(\mathbf{B}\), the corresponding calculation is

$$ \|\mathbf{B}\|_F =\sqrt{2^2+0^2+(-1)^2+1^2} =\sqrt{4+0+1+1} =\sqrt{6}. $$

Since \(3=\sqrt{9}>\sqrt{6}\), matrix \(\mathbf{A}\) has the larger Frobenius norm. Check: squaring the reported norms gives \(3^2=9\) and \((\sqrt{6})^2=6\), exactly the sums of squared entries calculated above.

The calculation treats every entry in the same way, regardless of its row or column. Therefore, rearranging the entries without changing any of their values does not change the sum of their squared absolute values. Transposition is one such rearrangement.

Theorem: If \(\mathbf{A}\in\mathbb{C}^{m\times n}\), then \(\|\mathbf{A}^{\top}\|_F=\|\mathbf{A}\|_F\). If \(c\in\mathbb{C}\), then \(\|c\mathbf{A}\|_F=|c|\|\mathbf{A}\|_F\). In particular, for real matrices and real scalars, the scalar factor is \(|c|\).

Proof. The entries of \(\mathbf{A}^{\top}\) are the entries of \(\mathbf{A}\) with their row and column positions exchanged. The double sum of their squared absolute values therefore has exactly the same terms: \(\|\mathbf{A}^{\top}\|_F^2=\sum_{i=1}^{n}\sum_{j=1}^{m}|a_{ji}|^2=\sum_{i=1}^{m}\sum_{j=1}^{n}|a_{ij}|^2=\|\mathbf{A}\|_F^2\). Both norms are nonnegative, so equality of their squares gives equality of the norms. For scalar multiplication, every entry of \(c\mathbf{A}\) is \(ca_{ij}\), so \(\|c\mathbf{A}\|_F^2=\sum_{i=1}^{m}\sum_{j=1}^{n}|ca_{ij}|^2=|c|^2\sum_{i=1}^{m}\sum_{j=1}^{n}|a_{ij}|^2=|c|^2\|\mathbf{A}\|_F^2\). Taking nonnegative square roots gives \(\|c\mathbf{A}\|_F=|c|\|\mathbf{A}\|_F\). This also covers \(c=0\), since both sides are then zero. \(\square\)

Worked Example: A Scalar Multiple and a Transpose

Let \(\mathbf{C}=\begin{bmatrix}1&2&-1\\0&-2&2\end{bmatrix}\). Find \(\|\mathbf{C}\|_F\), \(\|-\tfrac32\mathbf{C}\|_F\), and \(\|\mathbf{C}^{\top}\|_F\), computing each from its entries.

First, the sum of the squares of the entries of \(\mathbf{C}\) is \(1^2+2^2+(-1)^2+0^2+(-2)^2+2^2=1+4+1+0+4+4=14\). Thus \(\|\mathbf{C}\|_F=\sqrt{14}\). The scalar multiple is \(\begin{bmatrix}-\tfrac32&-3&\tfrac32\\0&3&-3\end{bmatrix}\), so direct calculation gives

$$ \left\|-\frac32\mathbf{C}\right\|_F^2 =\frac94+9+\frac94+0+9+9 =\frac{18}{4}+27 =\frac92+27 =\frac{63}{2}. $$

Therefore \(\|-\tfrac32\mathbf{C}\|_F=\sqrt{\tfrac{63}{2}}=\tfrac32\sqrt{14}\). The transpose is \(\mathbf{C}^{\top}=\begin{bmatrix}1&0\\2&-2\\-1&2\end{bmatrix}\), and its entries give \(\|\mathbf{C}^{\top}\|_F^2=1^2+0^2+2^2+(-2)^2+(-1)^2+2^2=1+0+4+4+1+4=14\), hence \(\|\mathbf{C}^{\top}\|_F=\sqrt{14}\).

Check: the scalar-multiple result agrees with \(|-\tfrac32|\|\mathbf{C}\|_F=\tfrac32\sqrt{14}\), whose square is \(\tfrac94\cdot14=\tfrac{63}{2}\). The transpose and original both have squared norm \(14\), as required.

What Happens to the Norm of a Sum?

For a sum, the squared norm is not generally the sum of the squared norms. The Frobenius inner product measures how the entries of the two matrices align: positive contributions can increase the squared norm of the sum, and negative contributions can decrease it. The following identity makes this exact. It applies to matrices of the same dimensions, so that both their sum and their inner product are defined.

Theorem: If \(\mathbf{A},\mathbf{B}\in\mathbb{C}^{m\times n}\), then \(\|\mathbf{A}+\mathbf{B}\|_F^2=\|\mathbf{A}\|_F^2+\|\mathbf{B}\|_F^2+2\operatorname{Re}\langle\mathbf{A},\mathbf{B}\rangle\). For real matrices, the inner product is real, so the last term is \(2\langle\mathbf{A},\mathbf{B}\rangle\).

Proof. By the definition of the induced norm and the linearity of the inner product in its first slot, \(\|\mathbf{A}+\mathbf{B}\|_F^2=\langle\mathbf{A}+\mathbf{B},\mathbf{A}+\mathbf{B}\rangle=\langle\mathbf{A},\mathbf{A}\rangle+\langle\mathbf{A},\mathbf{B}\rangle+\langle\mathbf{B},\mathbf{A}\rangle+\langle\mathbf{B},\mathbf{B}\rangle\). The inner product satisfies \(\langle\mathbf{B},\mathbf{A}\rangle=\overline{\langle\mathbf{A},\mathbf{B}\rangle}\). For any complex number \(z\), \(z+\overline z=2\operatorname{Re}z\). The two diagonal terms are \(\|\mathbf{A}\|_F^2\) and \(\|\mathbf{B}\|_F^2\), so collecting the cross terms proves the identity. Over \(\mathbb{R}\), taking the real part leaves the inner product unchanged. \(\square\)

Worked Example: The Sum Can Be Smaller Than One Summand

Let \(\mathbf{P}=\begin{bmatrix}1&-2\\2&0\end{bmatrix}\) and \(\mathbf{Q}=\begin{bmatrix}-1&1\\0&2\end{bmatrix}\). Calculate the norm of \(\mathbf{P}+\mathbf{Q}\), and use the identity to account for its value.

The individual squared norms are \(\|\mathbf{P}\|_F^2=1+4+4+0=9\) and \(\|\mathbf{Q}\|_F^2=1+1+0+4=6\). Their Frobenius inner product is the sum of the products of corresponding entries: \(\langle\mathbf{P},\mathbf{Q}\rangle=1(-1)+(-2)(1)+2(0)+0(2)=-1-2+0+0=-3\). The sum matrix is

$$ \mathbf{P}+\mathbf{Q} =\begin{bmatrix}1-1&-2+1\\2+0&0+2\end{bmatrix} =\begin{bmatrix}0&-1\\2&2\end{bmatrix}. $$

Calculating directly gives \(\|\mathbf{P}+\mathbf{Q}\|_F^2=0^2+(-1)^2+2^2+2^2=0+1+4+4=9\), so \(\|\mathbf{P}+\mathbf{Q}\|_F=3\). The identity gives the same result: \(\|\mathbf{P}+\mathbf{Q}\|_F^2=9+6+2(-3)=15-6=9\). Thus the sum has the same norm as \(\mathbf{P}\), even though \(\mathbf{Q}\) is nonzero. Check: the direct sum-of-squares calculation is \(9\), and the inner-product expansion is also \(9\).

Common Pitfall: Adding Norms as Though They Were Squares

A tempting but incorrect calculation is to claim that the norm of a sum always equals the sum of the norms, or that the squared norm always equals the sum of the squared norms. In the example above, the first claim would give \(\|\mathbf{P}+\mathbf{Q}\|_F=3+\sqrt{6}\), even though the entries of the sum give the value \(3\). The second claim would give \(9+6=15\) for the squared norm, instead of the correct value \(9\). Both errors ignore the cross term \(2\langle\mathbf{P},\mathbf{Q}\rangle=-6\).

To correct the calculation, first check the dimensions, then either add the matrices entry by entry and compute the norm from the resulting entries, or use the squared-norm identity and include the inner-product term. The two methods must agree. For a scalar multiple, use the absolute value of the scalar: multiplying a matrix by a negative real number still multiplies its norm by a nonnegative factor. For a transpose, the positions change but the entries do not, so the sum of their squared absolute values stays the same.

1
Check dimensions.
Matrices can be added only when they have the same numbers of rows and columns. A transpose exchanges the dimensions.
2
Use the entry formula.
Square the absolute value of every entry, add those values, and take the nonnegative square root.
3
Check the operation.
For scalar multiples, compare with \(|c|\|\mathbf{A}\|_F\); for transposes, compare with \(\|\mathbf{A}\|_F\); for sums, include the inner-product term.

These calculations give a consistent way to compare matrix sizes without confusing a matrix with any one of its entries. The entry formula is especially practical for exact examples, while the inner-product identity explains why adding matrices can either increase or decrease the resulting size.

Check Your Understanding

Show your entrywise calculations, and check each result against the relevant identity.

  1. Find the Frobenius norm of \(\begin{bmatrix}2&-1\\3&0\end{bmatrix}\).
  2. For \(\mathbf{A}=\begin{bmatrix}1&2\\-1&0\end{bmatrix}\), calculate \(\|2\mathbf{A}\|_F\) from its entries and compare it with \(2\|\mathbf{A}\|_F\).
  3. Calculate the Frobenius norm of \(\begin{bmatrix}1&0\\2&-1\\3&2\end{bmatrix}\) and of its transpose.
  4. For \(\mathbf{X}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\) and \(\mathbf{Y}=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\), find \(\|\mathbf{X}+\mathbf{Y}\|_F^2\) directly and by the sum identity.
  5. Identify the error in the claim \(\|\mathbf{A}+\mathbf{B}\|_F^2=\|\mathbf{A}\|_F^2+\|\mathbf{B}\|_F^2\) for all same-sized matrices.

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