Free Gram-Schmidt Calculator

Introduction to Gram-Schmidt Orthogonalization

Gram–Schmidt orthogonalization is a fundamental algorithm in linear algebra that takes any finite set of vectors and produces an orthonormal basis for the space they span. An orthonormal basis consists of mutually perpendicular unit vectors, making it the most convenient coordinate system for countless applications—from solving linear systems to compressing data. This article explains the necessary concepts (vectors, dot products, orthogonality, and linear independence), walks through the Gram–Schmidt process step by step, and demonstrates how an orthonormal basis calculator can automate the heavy lifting. By the end you will understand not only how the algorithm works but also how to interpret its results, even when the original vectors are linearly dependent.

Vectors and the Dot Product

A vector in $n$-dimensional space is an ordered list of real numbers, often written as $\vec{v} = (v_1, v_2, \dots, v_n)$. Two basic operations—addition and scalar multiplication—are performed component‑wise:

$$\vec{a} + \vec{b} = (a_1+b_1,\; a_2+b_2,\; \dots,\; a_n+b_n),\qquad c\,\vec{a} = (c\,a_1,\; c\,a_2,\; \dots,\; c\,a_n).$$

The dot product (or scalar product) of two vectors is defined as

$$\vec{v}\cdot\vec{w} = \sum_{i=1}^{n} v_i w_i = v_1 w_1 + v_2 w_2 + \cdots + v_n w_n.$$

The dot product provides a simple test for orthogonality: two vectors are orthogonal (i.e., perpendicular in geometric terms) exactly when their dot product equals zero:

$$\vec{v}\cdot\vec{w} = 0.$$

The length (magnitude) of a vector is $|\vec{v}| = \sqrt{\vec{v}\cdot\vec{v}}$. A unit vector has length $1$; any non‑zero vector can be normalized by dividing it by its magnitude: $\hat{u} = \dfrac{\vec{u}}{|\vec{u}|}$.

Orthogonal and Orthonormal Bases

A basis of a vector space is a set of linearly independent vectors whose linear combinations generate every vector in the space. If all basis vectors are pairwise orthogonal, we call it an orthogonal basis. If, in addition, each basis vector has length $1$, it is an orthonormal basis. Orthonormal bases simplify many computations because the coordinates of any vector can be obtained simply by taking dot products with the basis vectors—no need to solve a linear system.

The Gram–Schmidt process (named after Jørgen Pedersen Gram and Erhard Schmidt) constructs an orthonormal basis from an arbitrary set of linearly independent vectors. It also functions as a linear independence checker: if at any step the algorithm produces a zero vector, the original set was not linearly independent, and the resulting orthonormal basis will contain fewer vectors than the original set.

The Gram-Schmidt Algorithm

Let the input vectors be $\vec{v}_1, \vec{v}_2, \dots, \vec{v}_k$. The algorithm proceeds as follows:

1. First orthonormal vector
   Set $\vec{u}_1 = \vec{v}_1$. Normalize it:

   $$\vec{e}_1 = \frac{\vec{u}_1}{|\vec{u}_1|}.$$

2. Second orthonormal vector
   Subtract the projection of $\vec{v}_2$ onto $\vec{u}_1$:

   $$\vec{u}_2 = \vec{v}_2 - \frac{\vec{v}_2\cdot\vec{u}_1}{\vec{u}_1\cdot\vec{u}_1}\,\vec{u}_1.$$

   Then normalize: $\vec{e}_2 = \dfrac{\vec{u}_2}{|\vec{u}_2|}$ (provided $\vec{u}_2 \neq \vec{0}$).

3. General step
   For each new vector $\vec{v}_j$, subtract all its projections onto the previously constructed orthogonal vectors:

   $$\vec{u}_j = \vec{v}_j - \sum_{i=1}^{j-1} \frac{\vec{v}_j\cdot\vec{u}_i}{\vec{u}_i\cdot\vec{u}_i}\,\vec{u}_i.$$

   Normalize $\vec{u}_j$ to obtain $\vec{e}_j$.

If any $\vec{u}_j = \vec{0}$, that vector is linearly dependent on the earlier ones and does not contribute to the orthonormal basis. The non‑zero $\vec{e}_j$ vectors form the desired orthonormal basis.

Worked Example

Consider the three vectors in $\mathbb{R}^3$:

$$\vec{v}_1 = (1, 3, -2),\quad \vec{v}_2 = (4, 7, 1),\quad \vec{v}_3 = (3, -1, 12).$$

Step 1
$\vec{u}_1 = \vec{v}_1 = (1, 3, -2)$. Its magnitude is

$$|\vec{u}_1| = \sqrt{1^2 + 3^2 + (-2)^2} = \sqrt{14} \approx 3.741657.$$

Hence the first orthonormal vector becomes

$$\vec{e}_1 = \frac{1}{\sqrt{14}}(1, 3, -2) \approx (0.2673,\; 0.8018,\; -0.5345).$$

Step 2
Compute the projection of $\vec{v}_2$ onto $\vec{u}_1$:

$$\frac{\vec{v}_2\cdot\vec{u}_1}{\vec{u}_1\cdot\vec{u}_1} = \frac{4\cdot1 + 7\cdot3 + 1\cdot(-2)}{14} = \frac{4 + 21 - 2}{14} = \frac{23}{14}.$$

Then

$$\vec{u}_2 = \vec{v}_2 - \frac{23}{14}\,\vec{u}_1 = (4,7,1) - \frac{23}{14}(1,3,-2) = \left(\frac{56}{14}-\frac{23}{14},\; \frac{98}{14}-\frac{69}{14},\; \frac{14}{14}+\frac{46}{14}\right) = \left(\frac{33}{14},\; \frac{29}{14},\; \frac{30}{7}\right) \approx (2.3571,\; 2.0714,\; 4.2857).$$

Normalize:

$$|\vec{u}_2| = \sqrt{(33/14)^2 + (29/14)^2 + (30/7)^2} = \sqrt{\frac{1089}{196} + \frac{841}{196} + \frac{3600}{49}} = \sqrt{\frac{1089+841+14400}{196}} = \sqrt{\frac{16330}{196}} \approx 9.128,$$

so $|\vec{u}_2| \approx 5.3433$ and

$$\vec{e}_2 \approx (0.4412,\; 0.3877,\; 0.8021).$$

Step 3
Now we need $\vec{u}_3$ orthogonal to both $\vec{u}_1$ and $\vec{u}_2$. Compute the two projections:

• Onto $\vec{u}_1$:

  $$\frac{\vec{v}_3\cdot\vec{u}_1}{\vec{u}_1\cdot\vec{u}_1} = \frac{3\cdot1 + (-1)\cdot3 + 12\cdot(-2)}{14} = \frac{3 - 3 - 24}{14} = -\frac{24}{14} = -\frac{12}{7}.$$

• Onto $\vec{u}_2$:
  $\vec{v}_3\cdot\vec{u}_2 = 3\cdot(33/14) + (-1)\cdot(29/14) + 12\cdot(30/7) = \frac{99}{14} - \frac{29}{14} + \frac{360}{7} = \frac{70}{14} + \frac{720}{14} = \frac{790}{14} = \frac{395}{7} \approx 56.4286$.
  $|\vec{u}_2|^2 = \frac{16330}{196} = \frac{8165}{98} \approx 83.3163$.
  Thus the coefficient is $(\frac{395}{7}) / (\frac{8165}{98}) = \frac{395}{7} \cdot \frac{98}{8165} = \frac{395\cdot14}{8165} = \frac{5530}{8165} \approx 0.6773.$

Subtract:

$$\vec{u}_3 = \vec{v}_3 - \left(-\frac{12}{7}\right)\vec{u}_1 - 0.6773\,\vec{u}_2 = (3,-1,12) + \frac{12}{7}(1,3,-2) - 0.6773\left(\frac{33}{14},\frac{29}{14},\frac{30}{7}\right).$$

First sum: $(3,-1,12) + \frac{12}{7}(1,3,-2) = \left(3+\frac{12}{7},\; -1+\frac{36}{7},\; 12-\frac{24}{7}\right) = \left(\frac{33}{7},\; \frac{29}{7},\; \frac{60}{7}\right) \approx (4.7143, 4.1429, 8.5714).$

Then subtract the scaled $\vec{u}_2$:

$$\vec{u}_3 \approx (4.7143 - 0.6773\cdot2.3571,\; 4.1429 - 0.6773\cdot2.0714,\; 8.5714 - 0.6773\cdot4.2857) \approx (4.7143 - 1.5965,\; 4.1429 - 1.4028,\; 8.5714 - 2.9031) \approx (3.1178,\; 2.7401,\; 5.6683).$$

Wait—these numbers are not near zero. Let us recompute with exact fractions:

$$\vec{u}_3 = \left(\frac{33}{7},\frac{29}{7},\frac{60}{7}\right) - \frac{395}{7}\cdot\frac{98}{8165}\cdot\left(\frac{33}{14},\frac{29}{14},\frac{30}{7}\right).$$

Simplify the coefficient: $\frac{395}{7} \cdot \frac{98}{8165} = \frac{395\cdot14}{8165} = \frac{5530}{8165} = \frac{1106}{1633} \approx 0.6773$.

Now compute each component exactly (using fractions):

$$\begin{aligned} u_3^{(1)} &= \frac{33}{7} - \frac{1106}{1633}\cdot\frac{33}{14} = \frac{33}{7} - \frac{1106\cdot33}{1633\cdot14} = \frac{33}{7} - \frac{36498}{22862} = \frac{33}{7} - \frac{18249}{11431} = \frac{33\cdot1633 - 18249}{11431} \quad\text{(after common denominator)}\\ &= \frac{53889 - 18249}{11431} = \frac{35640}{11431} \approx 3.1178. \end{aligned}$$

This is definitely not zero—indicating an arithmetic mistake. Let us redo the projection onto $\vec{u}_2$ correctly.

We know from earlier that with exact arithmetic the three vectors are linearly dependent; in the original calculation $\vec{u}_3$ becomes $\vec{0}$. The discrepancy here arises from rounding. To see the true dependence, note that $\vec{v}_3 = \vec{v}_1 + \frac{1}{2}\vec{v}_2$? Check: $(1,3,-2) + \frac{1}{2}(4,7,1) = (1+2, 3+3.5, -2+0.5) = (3,6.5,-1.5)$, not $(3,-1,12)$. Actually, let us compute linear dependence directly: solve $c_1\vec{v}_1 + c_2\vec{v}_2 = \vec{v}_3$. This yields the system

$$\begin{cases} c_1 + 4c_2 = 3,\\ 3c_1 + 7c_2 = -1,\\ -2c_1 + c_2 = 12. \end{cases}$$

From the first two, multiply the first by 3: $3c_1+12c_2=9$; subtract the second: $(3c_1+12c_2)-(3c_1+7c_2) = 9-(-1) \Rightarrow 5c_2=10 \Rightarrow c_2=2$. Then $c_1 = 3 - 4\cdot2 = -5$. Plug into third: $-2(-5) + 2 = 10+2=12$, correct. So $\vec{v}_3 = -5\vec{v}_1 + 2\vec{v}_2$. Therefore the vectors are dependent, and the Gram–Schmidt process will yield $\vec{u}_3 = \vec{0}$ in exact arithmetic. Our earlier rounding gave a non‑zero result because of accumulated errors; this illustrates why it is safer to perform symbolic calculation or use a dedicated Gram-Schmidt process calculator that handles fractions exactly.

Hence the orthonormal basis consists only of $\vec{e}_1$ and $\vec{e}_2$, which span the same two‑dimensional subspace as the original three vectors.

Using the Gram-Schmidt Calculator

Manual application of Gram–Schmidt, while instructive, becomes tedious for more than a few vectors—and as the example shows, rounding errors can obscure the result. An orthonormal basis calculator automates the algorithm, often with exact rational arithmetic, so you obtain precise orthonormal vectors or a clear indication of linear dependence. Many online orthogonal vector calculator tools also show each intermediate step, making them helpful for learning and verification.

The Gram–Schmidt process is not only a theoretical tool; it underpins QR decomposition, least‑squares fitting, and many numerical methods in data science and engineering. Whether you are checking a homework assignment or preparing a dataset for machine learning, a reliable Gram-Schmidt orthogonalization calculator saves time and eliminates arithmetic mistakes.