Lesson Explainer: Orthogonal Matrices | Nagwa Lesson Explainer: Orthogonal Matrices | Nagwa

Lesson Explainer: Orthogonal Matrices Mathematics

In this explainer, we will learn how to determine whether a matrix is orthogonal and how to find its inverse if it is.

In linear algebra, there are many special types of matrices that are interesting either because of the geometric transformations that they represent, or because of the convenient algebraic properties that they hold. Very often, these special matrices are square matrices and have a definition that is in some way related to the determinant or the transpose. For example, symmetric matrices are square matrices which are equal to their own transpose, and skew-symmetric matrices are equal to the their own transpose after a sign change in every entry. These types of special matrices have a plethora of applications and are full of a range of algebraic properties which make them very attractive in a theoretical sense. For this explainer, we will be interested in orthogonal matrices, which have a very particular and restrictive definition. Orthogonal matrices are defined by two key concepts in linear algebra: the transpose of a matrix and the inverse of a matrix. Orthogonal matrices also have a deceptively simple definition, which gives a helpful starting point for understanding their general algebraic properties.

Definition: Orthogonal Matrix

For a square matrix 𝐴 to be orthogonal, it must be the case that 𝐴𝐴=𝐼, where 𝐴 is the matrix transpose of 𝐴 and where 𝐼 is the 𝑛×𝑛 identity matrix.

If we were to take a random square matrix, then it is very unlikely that this matrix would also be orthogonal. To demonstrate this, take the following square matrix where the entries are random integers: 𝐴=11243136613.

The transpose of 𝐴 is 𝐴=14613621313.

To check if 𝐴 is orthogonal, we need to see whether 𝐴𝐴=𝐼, where 𝐼 is the 3×3 identity matrix 𝐼=100010001.

Using matrix multiplication, we would find that 𝐴𝐴=1124313661314613621313=633383319421138211241.

Clearly, it is absolutely not the case that 𝐴𝐴=𝐼, and therefore 𝐴 is not an orthogonal matrix.

At this stage, it might become apparent that it is unlikely that a random square matrix would be orthogonal. As we will see later, there are very strong conditions which are necessary for a matrix to be orthogonal, and these can, to some extent, be thought of algebraically. We will first consider two examples as a way of practicing our ability to determine whether or not a given matrix is orthogonal.

Example 1: Determining Whether a 3 × 3 Matrix is Orthogonal

Is the matrix 𝐴=13122212221 orthogonal?


For the matrix 𝐴 to be orthogonal, it must be the case that 𝐴𝐴=𝐼, where 𝐼 is the 3×3 identity matrix. Given that 𝐴=13122212221, we can multiply these two matrices together to give 𝐴𝐴=13×13122212221122212221=19900090009=100010001=𝐼.

Since we have found that 𝐴𝐴=𝐼, it is the case that 𝐴 is orthogonal.

Example 2: Determining Whether a 3 × 3 Matrix is Orthogonal

Is the matrix 𝐴=122212221 orthogonal?


If 𝐴 is orthogonal, then 𝐴𝐴=𝐼, where 𝐼 is the 3×3 identity matrix, sometimes referred to as the 3×3 unit matrix. To check for orthogonality, we can find the transpose matrix 𝐴=122212221.

Then, we perform the matrix multiplication 𝐴𝐴=122212221122212221=988898889.

Given that 𝐴𝐴𝐼, the matrix 𝐴 is not orthogonal.

Provided that we have a good understanding of matrix multiplication, it is straightforward to verify whether a given matrix is orthogonal, although we will have to perform many calculations to complete the matrix multiplication for matrices with larger orders. We might then reasonably ask if there are any other methods for determining whether or not a matrix is orthogonal. Since any orthogonal matrix must be a square matrix, we might expect that we can use the determinant to help us in this regard, given that the determinant is only defined for square matrices. The determinant is a concept that has a range of very helpful properties, several of which contribute to the proof of the following theorem.

Theorem: Determinant of an Orthogonal Matrix

Supposing that 𝐴 is an orthogonal matrix, then it must be the case that the determinant of 𝐴 can take only two values. Specifically, it must be the case that |𝐴|=±1.

We recall the definition of an orthogonal matrix, which states that for 𝐴 to be orthogonal, it must be that 𝐴𝐴=𝐼.

By taking determinants of both sides, we obtain |𝐴𝐴|=|𝐼|.

The determinant is multiplicative over matrix multiplication, which means that |𝐵𝐶|=|𝐵||𝐶| for any two square matrices of equal dimension, 𝐵 and 𝐶. The above equation then becomes |𝐴||𝐴|=|𝐼|.

Two standard results from linear algebra are that the determinant of a transpose matrix is equal to the determinant of the original matrix; in other words, |𝐴|=|𝐴|. It is also true that the determinant of any identity matrix is equal to 1, meaning that |𝐼|=1. Applying both of these results to the above equation gives |𝐴||𝐴|=1.

We, therefore, have that |𝐴|=1, which shows that det(𝐴)=±1, as required.

We now know that, for a square matrix 𝐴 to be orthogonal, it is necessary for that matrix to have a determinant of ±1. However, that is not in itself a sufficient condition for orthogonality. As an example, suppose we take the matrix 𝐴=||||111326223||||.

Since 𝐴 is a 3×3 matrix, we can use Sarrus’ rule to calculate the determinant as follows: |𝐴|=𝑎|𝐴|𝑎|𝐴|+𝑎|𝐴|=1×||2623||(1)×||3623||+(1)×||3222||=1×6(1)×(3)+(1)×(2)=1.

The determinant of 𝐴 is equal to 1; therefore, it is possible that the matrix is orthogonal. We test this by constructing the transpose matrix 𝐴=132122163 and then performing the calculation 𝐴𝐴=111326223132122163=311711492872817.

Given that 𝐴𝐴𝐼, the matrix is not orthogonal despite the fact that it has a determinant of 1.

The above theorem is helpful as it can tell us immediately whether it is possible for a square matrix to be orthogonal. Whilst the theorem does give us a necessary condition for orthogonality, it is not in itself a sufficient condition for orthogonality, as we saw in the previous example. Practically, though, it is generally wise to calculate the determinant of a square matrix before checking whether it is orthogonal. If a square matrix has a determinant of ±1, then it is possible that it will be an orthogonal matrix, although determining this with certainty will require the following definition and theorem.

Definition: Dot Product of Two Vectors

Consider the two vectors 𝑢=(𝑢𝑢𝑢),𝑣=(𝑣𝑣𝑣). Then the dot product, also known as the scalar product, of these two vectors is defined by the formula 𝑢𝑣=𝑢×𝑣=(𝑢×𝑣)+(𝑢×𝑣)++(𝑢×𝑣).

Theorem: Orthogonal Matrices and Relationships between Columns

Suppose that we have the square matrix 𝐴=𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 and that the columns of this matrix are labelled as 𝑐=(𝑎𝑎𝑎),𝑐=(𝑎𝑎𝑎),𝑐=(𝑎𝑎𝑎).

Then, for 𝐴 to be orthogonal, it must be the case that 𝑐𝑐=1 for all 𝑖 and that 𝑐𝑐=0 for any 𝑖𝑗, where indicates the dot product. If the columns of 𝐴 have this property, then they are called an orthonormal set.

This gives us a test by which we can diagnose whether or not a matrix is orthogonal. In the following example, we will apply the test described in the theorem above; however, we will first check whether the determinant is equal to ±1, as otherwise it will not be possible for the matrix to be orthogonal. Although it is not strictly necessary to perform this step, neglecting to do so might mean a lot of wasted effort if the matrix is not actually orthogonal.

Example 3: Determining Whether a 2 × 2 Matrix is Orthogonal by Using the Dot Product

Determine whether the following matrix is orthogonal: 𝐴=32121232.


For a square matrix to be orthogonal, it must have a determinant equal to ±1. For the matrix 𝐴, we can use the well-known formula for the determinant of a 2×2 matrix: 𝑎𝑏𝑐𝑑=𝑎𝑑𝑏𝑐.

Applying this result to the given matrix 𝐴, we have |𝐴|=||||||32121232||||||=32×3212×12=1.

It is possible that this matrix is orthogonal, but to know this for certain we will have to compare the columns. We can separately write out the two columns of the matrix as the vectors 𝑐=3212,𝑐=1232.

For the matrix 𝐴 to be orthogonal, the above column vectors must have some special properties. First, it must be the case that 𝑐𝑐=1 for 𝑖=1,2. It must also be the case that 𝑐𝑐=0 when 𝑖𝑗. In other words, we must check that 𝑐𝑐=1, 𝑐𝑐=1, and 𝑐𝑐=0.

We first highlight the entries of each column vector as shown: 𝑐=3212,𝑐=1232.

Then we have 𝑐𝑐=32×32+12×12=1, and 𝑐𝑐=12×12+32×32=1, as required. There is now only one condition remaining to check, so we calculate 𝑐𝑐=32×12+12×32=0.

The three stated conditions have been satisfied, and therefore 𝐴 is an orthogonal matrix, which we could check by seeing that it meets the definition and obeys the property 𝐴𝐴=𝐼.

In the above example, we have applied a theorem to check whether the given matrix was orthogonal. Another way of interpreting this theorem would be that, given a partially populated matrix, we now know how to populate the blank entries in a way that forces the matrix to be orthogonal. Depending on the values of the entries which have already been populated, it may not be possible to populate the blank entries in a way which forces the matrix to be orthogonal, although in the following example the known entries have been chosen in a way that will still allow this.

Example 4: Determining Whether a 3 × 3 Matrix Is Orthogonal by Using the Dot Product

Given that the matrix 23222623𝑎𝑏𝑐0𝑑 is orthogonal, find the values of 𝑎, 𝑏, 𝑐, and 𝑑.


We first label the above matrix as 𝐴 and write the three column vectors 𝑐=2323𝑐,𝑐=22𝑎0𝑐=26𝑏𝑑.

If 𝐴 is orthogonal, then it must be the case that 𝑐𝑐=1 for all 𝑖=1,2,3 and that 𝑐𝑐=0 for 𝑖𝑗, where the symbol represents the dot product between the two vectors. In other words, we must have 𝑐𝑐=1, 𝑐𝑐=1, and 𝑐𝑐=1. Additionally, we require that 𝑐𝑐=0, 𝑐𝑐=0, and 𝑐𝑐=0.

These relationships must all hold since 𝐴 is orthogonal, so we can use any of these relationships to help us determine the unknown variables 𝑎, 𝑏, 𝑐, and 𝑑. We will begin by using 𝑐 and 𝑐 to find the parameter 𝑎, by using the relationship 𝑐𝑐=0. Writing this out in full, we have 𝑐𝑐=23×22+23×𝑎+(𝑐×0)=23+23𝑎.

Given that 𝑐𝑐=0, we conclude that 𝑎=22. This means that we now have the three column vectors 𝑐=2323𝑐,𝑐=22220,𝑐=26𝑏𝑑.

The parameter 𝑏 can be found in a similar manner by using 𝑐 and 𝑐. We take the dot product 𝑐𝑐=22×26+22×𝑏+(0×𝑑)=1622𝑏.

We require that 𝑐𝑐=0, which implies that 𝑏=26. This value can be replaced in the column vectors, giving the updated versions 𝑐=2323𝑐,𝑐=22220,𝑐=2626𝑑.

If we were to now use the column vectors 𝑐 and 𝑐 and take the dot product between them, then we will involve the parameters 𝑐 and 𝑑. We find 𝑐𝑐=23×26+23×26+(𝑐×𝑑)=229+𝑐𝑑.

We must have 𝑐𝑐=0, which means that


This has not quite solved the problem yet, as it has only expressed the parameters 𝑐 and 𝑑 in a way that they are multiplied together. However, this does not mean that the result cannot be useful to us. We can already deduce that 𝑐 and 𝑑 must have opposite signs, which we will need to remember when it comes to stating the final result. We can also use the expressions that we derived earlier which involved taking the dot product of a column vector with itself. For example, to find 𝑐, we can now use the restriction 𝑐𝑐=1. By taking the dot product of 𝑐 with itself, we find 𝑐𝑐=23×23+23×23+(𝑐×𝑐)=89+𝑐.

Given that 𝑐𝑐=1, we have 𝑐=19, and hence 𝑐=±13. By using equation (1), we can find 𝑑 by rearranging to give 𝑑=1𝑐×229=(±3)×229=223.

As an additional check that 𝑑 is the correct value, we could verify that 𝑐𝑐=1 with either possible value of 𝑑. We reasoned earlier that 𝑐 and 𝑑 have opposite signs, so we can conclude that there are two possible values for 𝑐 and 𝑑. Therefore, there are two possible versions of the correct column vectors 𝑐=2323±13,𝑐=22220,𝑐=2626223.

Accordingly, there are two possible forms for the matrix 𝐴 to be orthogonal, which are summarized by the single expression 𝐴=232226232226±130223.

As a check that this matrix is definitely an orthogonal matrix, we could check that 𝐴𝐴=𝐼, where 𝐼 is the 3×3 identity matrix.

We now have a total of three tests to decide whether or not a matrix is orthogonal. For a matrix 𝐴, we can first check whether orthogonality is even possible by seeing if |𝐴|=±1. If orthogonality is possible, then we can see whether 𝐴𝐴=𝐼, which only requires one instance of matrix multiplication. Separate from these two methods, we can also compare the columns of 𝐴 and see whether they form an orthonormal set. Whilst these tests are interesting, they are not overtly helpful if we are interested in constructing an orthogonal matrix. Unsurprisingly, there is an algorithm for creating an orthogonal matrix from a set of starting vectors, which is referred to as the Gram–Schmidt algorithm. This algorithm is generally considered to be one of the most useful algorithms in all of linear algebra, as orthonormal sets are the foundation of many modern fields such as computer visualization and quantum field theory.

Orthogonal matrices are also considered to be especially important because of their relationship to reflections and rotations in geometry. They also have the highly convenient property that their transpose is equal to their own inverse, which can be easily deduced from the definition. Suppose we assume that 𝐴 is an orthogonal square matrix, which means that 𝐴𝐴=𝐼.

Since 𝐴 is orthogonal, we know that the determinant is equal to ±1. Given that the determinant is nonzero, this means that 𝐴 is invertible, and hence 𝐴 exists. If we now multiply the above equation on the left-hand side by 𝐴, we find 𝐴𝐴𝐴=𝐴𝐼.

We know that matrix multiplication is associative, which means that 𝐵(𝐶𝐷)=(𝐵𝐶)𝐷 for any matrices 𝐵, 𝐶, and 𝐷 which have compatible orders. This result means that we may write the above equation as 𝐴𝐴𝐴=𝐴𝐼.

By definition, we have 𝐴𝐴=𝐼, where 𝐼 is the 𝑛×𝑛 identity matrix. This gives 𝐼𝐴=𝐴𝐼.

The final result we need is the key property of the identity matrix that 𝐵𝐼=𝐼𝐵=𝐵 whenever 𝐵 is a matrix with suitable order. The above result, then, simplifies to the final form 𝐴=𝐴.

This is a key, defining feature of orthogonal matrices. Normally we would expect that the transpose of a matrix 𝐴 would be much easier to calculate than the multiplicative inverse 𝐴, which is typically a long-winded and error-prone process. But for orthogonal matrices the transpose is actually equal to the multiplicative inverse, which is a blessing should we ever wish to make use of the inverse matrix (as we most frequently do).

Key Points

  • A square matrix 𝐴 is orthogonal if 𝐴𝐴=𝐼, where 𝐼 is the 𝑛×𝑛 identity matrix.
  • For a matrix 𝐴 to be orthogonal, it must be the case that |𝐴|=±1.
  • Suppose that 𝑐,𝑐,,𝑐 represent the columns of a square matrix 𝐴. If 𝑐𝑐=1 for all 𝑖=1,2,,𝑛 and 𝑐𝑐=0 for all 𝑖𝑗, then the matrix 𝐴 is orthogonal.

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